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The meaning of "P is monotove relative to set containment" is this  : If $A\subset B$ then $P(A)\leq P(B)$. Now here for every $\omega$, your $A=\{\omega: |(X_n(\omega)+Y_n(\omega))-(X(\omega)+Y(\omega))|\geq \epsilon\}$ and your $B=\{\omega:|X_n(\omega)-X(\omega)|+|Y_n(\omega)-Y(\omega)|\geq \epsilon\}$. From your first in-equality and for every $\omega$ and $\epsilon >0$, we have: $\{|(X_n(\omega)+Y_n(\omega)-(X(\omega)+Y(\omega))|\geq \epsilon\} \subset \{|X_n(\omega)-X(\omega)|+|Y_n(\omega)-Y(\omega)|\geq \epsilon\}$.
So by above argument $P[|(X_n(\omega)+Y_n(\omega))-(X(\omega)+Y(\omega))|\geq \epsilon]\leq P[|X_n(\omega)-X(\omega)|+|Y_n(\omega)-Y(\omega)|\geq \epsilon].$ To show the last in-equlity, first we prove that for every for every $\omega$ and $\epsilon >0$ we have : $\{\omega:|X_n(\omega)-X(\omega)|+|Y_n(\omega)-Y(\omega)|\geq \epsilon\}\subset \Big\{\{\omega:|X_n(\omega)-X(\omega)|\geq \epsilon/2\}\cup\{\omega:|Y_n(\omega)-Y(\omega)|\geq \epsilon/2\}\Big\}.$
This can be simply shown by contradiction, i.e. if $\{\omega:|X_n(\omega)-X(\omega)|<\epsilon/2\}$ and $\{\omega:|Y_n(\omega)-Y(\omega)|<\epsilon/2\}$ then we can sum them up to get: $\{\omega:|X_n(\omega)-X(\omega)|+|Y_n(\omega)-Y(\omega)|<\epsilon\}$, which is a cotradiction.
Hence,
$P[\{\omega:|X_n(\omega)-X(\omega)|+|Y_n(\omega)-Y(\omega)|\geq \epsilon\}]\leq P[\big\{\{\omega:|X_n(\omega)-X(\omega)|\geq \epsilon/2\}\cup\{\omega:|Y_n(\omega)-Y(\omega)|\geq \epsilon/2\}\big\}]\leq P[\{\omega:|X_n(\omega)-X(\omega)|\geq \epsilon/2\}]+P[\{\omega:|Y_n(\omega)-Y(\omega)|\geq \epsilon/2\}],$
where the last in-equality comes from $P[C\cup D]\leq P[C]+P[D]$. Cheers!

The meaning of $P$ is monotone relative to set containment is this:

If $A \subseteq B$ then $P(A)\leq P(B)$.

Now here for every $\omega\in\Omega$, we have $$\begin{align*} A &= \left\{\omega : \left| (X_n(\omega) + Y_n(\omega)) - ( X(\omega) + Y(\omega) ) \right| \geq \epsilon \right\}, \text{ and}\\ B &= \{ \omega : \left| X_n(\omega) - X(\omega) \right| + \left| Y_n(\omega) - Y(\omega) \right| \geq \epsilon \}. \end{align*}$$

From your first inequality and for every $\omega\in\Omega$ and $\varepsilon >0$, we have

$$\left\{ \left| (X_n(\omega) + Y_n(\omega) - (X(\omega) + Y(\omega)) \right| \geq \varepsilon \right\} \subseteq \left\{ \left|X_n(\omega) - X(\omega)\right| + \left|Y_n(\omega) - Y(\omega)\right| \geq \varepsilon \right\}.$$

So from the statement above

$$P\left( \left| (X_n(\omega) + Y_n(\omega) - (X(\omega) + Y(\omega)) \right| \geq \varepsilon \right) \leq P\left( \left|X_n(\omega) - X(\omega)\right| + \left|Y_n(\omega) - Y(\omega)\right| \geq \varepsilon \right).$$

To show the last inequality, first we prove that for every for every $\omega\in\Omega$ and $\varepsilon >0$ we have

$$\begin{multline*} \{\omega : \left|X_n(\omega) - X(\omega)\right| + \left|Y_n(\omega) - Y(\omega)\right| \geq \varepsilon \}\\ \subseteq \{\omega : \left|X_n(\omega) - X(\omega) \right| \geq \varepsilon/2\} \cup \{\omega : \left|Y_n(\omega) - Y(\omega) \right| \geq \varepsilon/2\}. \end{multline*}$$

This can be simply shown by contradiction, i.e. if

$$\{ \omega : \left|X_n(\omega) - X(\omega) \right| < \varepsilon/2 \} \text{ and } \{\omega : \left|Y_n(\omega) - Y(\omega) \right| < \varepsilon/2 \}$$

then we can sum them up to get $\{ \omega : \left|X_n(\omega) - X(\omega)\right| + \left|Y_n(\omega) - Y(\omega)\right| < \varepsilon \}$, which is a contradiction.

Hence,
$$\begin{align*} P\bigl( \{ \omega : &\left| X_n(\omega)-X(\omega) \right| + \left| Y_n(\omega) - Y(\omega) \right|\geq \varepsilon \} \bigr)\\ &\leq P\left( \{\omega : \left| X_n(\omega) - X(\omega) \right| \geq \varepsilon/2\} \cup \{\omega : \left| Y_n(\omega) - Y(\omega) \right| \geq \varepsilon/2 \} \right)\\ &\leq P\left( \{\omega : \left| X_n(\omega) - X(\omega) \right| \geq \varepsilon/2\}\right) + P\left( \{\omega : \left| Y_n(\omega) - Y(\omega) \right| \geq \varepsilon/2 \} \right), \end{align*}$$
where the last inequality comes from $P(C \cup D) \leq P(C) + P(D)$. Cheers!

The meaning of "P is monotove relative to set containment" is this : If $A\subset B$ then $P(A)\leq P(B)$. Now here for every $\omega$, your $A=\{\omega: |(X_n(\omega)+Y_n(\omega))-(X(\omega)+Y(\omega))|\geq \epsilon\}$ and your $B=\{\omega:|X_n(\omega)-X(\omega)|+|Y_n(\omega)-Y(\omega)|\geq \epsilon\}$. From your first in-equality and for every $\omega$ and $\epsilon >0$, we have: $\{|(X_n(\omega)+Y_n(\omega)-(X(\omega)+Y(\omega))|\geq \epsilon\} \subset \{|X_n(\omega)-X(\omega)|+|Y_n(\omega)-Y(\omega)|\geq \epsilon\}$.
So by above argument $P[|(X_n(\omega)+Y_n(\omega))-(X(\omega)+Y(\omega))|\geq \epsilon]\leq P[|X_n(\omega)-X(\omega)|+|Y_n(\omega)-Y(\omega)|\geq \epsilon].$ To show the last in-equlity, first we prove that for every for every $\omega$ and $\epsilon >0$ we have : $\{\omega:|X_n(\omega)-X(\omega)|+|Y_n(\omega)-Y(\omega)|\geq \epsilon\}\subset \Big\{\{\omega:|X_n(\omega)-X(\omega)|\geq \epsilon/2\}\cup\{\omega:|Y_n(\omega)-Y(\omega)|\geq \epsilon/2\}\Big\}.$
This can be simply shown by contradiction, i.e. if $\{\omega:|X_n(\omega)-X(\omega)|<\epsilon/2\}$ and $\{\omega:|Y_n(\omega)-Y(\omega)|<\epsilon/2\}$ then we can sum them up to get: $\{\omega:|X_n(\omega)-X(\omega)|+|Y_n(\omega)-Y(\omega)|<\epsilon\}$, which is a cotradiction.
Henec,
$P[\{\omega:|X_n(\omega)-X(\omega)|+|Y_n(\omega)-Y(\omega)|\geq \epsilon\}]\leq P[\big\{\{\omega:|X_n(\omega)-X(\omega)|\geq \epsilon/2\}\cup\{\omega:|Y_n(\omega)-Y(\omega)|\geq \epsilon/2\}\big\}]\leq P[\{\omega:|X_n(\omega)-X(\omega)|\geq \epsilon/2\}]+P[\{\omega:|Y_n(\omega)-Y(\omega)|\geq \epsilon/2\}],$
where the last in-equality comes from $P[C\cup D]\leq P[C]+P[D]$. Cheers!

The meaning of "P is monotove relative to set containment" is this : If $A\subset B$ then $P(A)\leq P(B)$. Now here for every $\omega$, your $A=\{\omega: |(X_n(\omega)+Y_n(\omega))-(X(\omega)+Y(\omega))|\geq \epsilon\}$ and your $B=\{\omega:|X_n(\omega)-X(\omega)|+|Y_n(\omega)-Y(\omega)|\geq \epsilon\}$. From your first in-equality and for every $\omega$ and $\epsilon >0$, we have: $\{|(X_n(\omega)+Y_n(\omega)-(X(\omega)+Y(\omega))|\geq \epsilon\} \subset \{|X_n(\omega)-X(\omega)|+|Y_n(\omega)-Y(\omega)|\geq \epsilon\}$.
So by above argument $P[|(X_n(\omega)+Y_n(\omega))-(X(\omega)+Y(\omega))|\geq \epsilon]\leq P[|X_n(\omega)-X(\omega)|+|Y_n(\omega)-Y(\omega)|\geq \epsilon].$ To show the last in-equlity, first we prove that for every for every $\omega$ and $\epsilon >0$ we have : $\{\omega:|X_n(\omega)-X(\omega)|+|Y_n(\omega)-Y(\omega)|\geq \epsilon\}\subset \Big\{\{\omega:|X_n(\omega)-X(\omega)|\geq \epsilon/2\}\cup\{\omega:|Y_n(\omega)-Y(\omega)|\geq \epsilon/2\}\Big\}.$
This can be simply shown by contradiction, i.e. if $\{\omega:|X_n(\omega)-X(\omega)|<\epsilon/2\}$ and $\{\omega:|Y_n(\omega)-Y(\omega)|<\epsilon/2\}$ then we can sum them up to get: $\{\omega:|X_n(\omega)-X(\omega)|+|Y_n(\omega)-Y(\omega)|<\epsilon\}$, which is a cotradiction.
Hence,
$P[\{\omega:|X_n(\omega)-X(\omega)|+|Y_n(\omega)-Y(\omega)|\geq \epsilon\}]\leq P[\big\{\{\omega:|X_n(\omega)-X(\omega)|\geq \epsilon/2\}\cup\{\omega:|Y_n(\omega)-Y(\omega)|\geq \epsilon/2\}\big\}]\leq P[\{\omega:|X_n(\omega)-X(\omega)|\geq \epsilon/2\}]+P[\{\omega:|Y_n(\omega)-Y(\omega)|\geq \epsilon/2\}],$
where the last in-equality comes from $P[C\cup D]\leq P[C]+P[D]$. Cheers!

The meaning of "P is monotove relative to set containment" is this : If $A\subset B$ then $P(A)\leq P(B)$. Now here for every $\omega$, your $A=\{\omega: |(X_n(\omega)+Y_n(\omega))-(X(\omega)+Y(\omega))|\geq \epsilon\}$ and your $B=\{\omega:|X_n(\omega)-X(\omega)|+|Y_n(\omega)-Y(\omega)|\geq \epsilon\}$. From your first in-equality and for every $\omega$ and $\epsilon >0$, we have: $\{|(X_n(\omega)+Y_n(\omega)-(X(\omega)+Y(\omega))|\geq \epsilon\} \subset \{|X_n(\omega)-X(\omega)|+|Y_n(\omega)-Y(\omega)|\geq \epsilon\}$.
So by above argument $P[|(X_n(\omega)+Y_n(\omega))-(X(\omega)+Y(\omega))|\geq \epsilon]\leq P[|X_n(\omega)-X(\omega)|+|Y_n(\omega)-Y(\omega)|\geq \epsilon].$

The meaning of "P is monotove relative to set containment" is this : If $A\subset B$ then $P(A)\leq P(B)$. Now here for every $\omega$, your $A=\{\omega: |(X_n(\omega)+Y_n(\omega))-(X(\omega)+Y(\omega))|\geq \epsilon\}$ and your $B=\{\omega:|X_n(\omega)-X(\omega)|+|Y_n(\omega)-Y(\omega)|\geq \epsilon\}$. From your first in-equality and for every $\omega$ and $\epsilon >0$, we have: $\{|(X_n(\omega)+Y_n(\omega)-(X(\omega)+Y(\omega))|\geq \epsilon\} \subset \{|X_n(\omega)-X(\omega)|+|Y_n(\omega)-Y(\omega)|\geq \epsilon\}$.
So by above argument $P[|(X_n(\omega)+Y_n(\omega))-(X(\omega)+Y(\omega))|\geq \epsilon]\leq P[|X_n(\omega)-X(\omega)|+|Y_n(\omega)-Y(\omega)|\geq \epsilon].$ To show the last in-equlity, first we prove that for every for every $\omega$ and $\epsilon >0$ we have : $\{\omega:|X_n(\omega)-X(\omega)|+|Y_n(\omega)-Y(\omega)|\geq \epsilon\}\subset \Big\{\{\omega:|X_n(\omega)-X(\omega)|\geq \epsilon/2\}\cup\{\omega:|Y_n(\omega)-Y(\omega)|\geq \epsilon/2\}\Big\}.$
This can be simply shown by contradiction, i.e. if $\{\omega:|X_n(\omega)-X(\omega)|<\epsilon/2\}$ and $\{\omega:|Y_n(\omega)-Y(\omega)|<\epsilon/2\}$ then we can sum them up to get: $\{\omega:|X_n(\omega)-X(\omega)|+|Y_n(\omega)-Y(\omega)|<\epsilon\}$, which is a cotradiction.
Henec,
$P[\{\omega:|X_n(\omega)-X(\omega)|+|Y_n(\omega)-Y(\omega)|\geq \epsilon\}]\leq P[\big\{\{\omega:|X_n(\omega)-X(\omega)|\geq \epsilon/2\}\cup\{\omega:|Y_n(\omega)-Y(\omega)|\geq \epsilon/2\}\big\}]\leq P[\{\omega:|X_n(\omega)-X(\omega)|\geq \epsilon/2\}]+P[\{\omega:|Y_n(\omega)-Y(\omega)|\geq \epsilon/2\}],$
where the last in-equality comes from $P[C\cup D]\leq P[C]+P[D]$. Cheers!