Probability of absolute value of a sum of random variables Consider two random variables $X$ and $Y$, and let $b$ be a real number. Show that
$$P(\mid X+Y\mid>b)\leq P(\mid X+Y\mid>b,\mid X\mid>b/2)+P(\mid X+Y\mid>b,\mid Y\mid>b/2).$$
I'm unable to understand/show this seemly basic fact.

UPDATE
The aim of this question was primarly to prove that
$$P(\mid X+Y\mid>b)\leq P(\mid X\mid>b/2)+P(\mid Y\mid>b/2).$$
If you can prove this rigorously, please answer. For me, it is sufficient to prove that 
$$\{w:\mid X \mid(w)+\mid Y \mid(w)>b\}\subseteq \{w:\mid X \mid(w)>b/2\}\cup\{w:\mid Y \mid(w)>b/2\}$$
 A: You can find the solution by drawing this on a X-Y plane. Once you see what's going, it's easy to come to a rigorous answer.
The left hand side of your original problem. On X-Y plane the set $|X+Y|>b$ is area outside two inclined lines as shown in picture below:

The right hand side of your original problem has two parts. First, look at the set $|X|>b/2$ in green color outside the two vertical lines:

Next, look at the set $|Y|>b/2$ in blue color outside the two horizontal lines:

We can combine last two plot and get the union of sets $|X|>b/2$ and $|Y|>b/2$, i.e. all green abd blue (or both) areas:

The union of sets $|X|>b/2$ and $|Y|>b/2$ includes  the set $|X+Y|>b$: everywhere where the red color shows up you have either green or blue (or both) colors too.
Therefore the measure must be smaller:$$P(|X+Y|>b)\le P(|X|>b/2)+P(|Y|>b/2)$$
This also shows that your question as it is posed is incorrect: $P(|X+Y|>b)\le P(|X+Y|>b,|X|>b/2)+P(|X+Y|>b,|Y|>b/2)$
A: 
Proof of $P(|X+Y|>b)\leq P(|X|>b/2)+P(|Y|>b/2)$ which is what the OP's  edited question says he really wanted to prove in the first place...

The event $\left\{|X|\leq \frac b2, |Y| \leq \frac b2\right\}$, that is, the event that $(X,Y)$ lies in the square of side $b$ centered at the origin, is clearly a subset of the event $\{|X+Y|\leq b\}$, and hence, taking complements of both (and using DeMorgan's laws), we have that the event $\left\{|X|> \frac b2\right\} \cup  \left\{|Y| > \frac b2\right\} = A \cup B$ is a superset of the event $\{|X+Y|> b\} =C$. Hence, \begin{align}
P\bigg\{|X+Y|> b\bigg\} &\leq P\left(\left\{|X|> \frac b2\right\} \cup  \left\{|Y| > \frac b2\right\}\right)\\ &\leq P\left\{|X|> \frac b2\right\} + P\left\{|Y|> \frac b2\right\}\end{align}
where the second step is just using the fact that $P(A\cup B) \leq P(A)+P(B)$ which is sometimes referred to as the simplest form of Boole's inequality.

Non-proof of original claim

It was shown above that $C \subset A\cup B$ and so it just be that 
\begin{align}
C &\subset A\cup B\\
&\Downarrow\\
C &\subset (C\cap A) \cup (C \cap B)\\
&\Downarrow\\
P(C) &\leq P\big((C\cap A) \cup (C \cap B)\big)\\
&\leq P(C\cap A)+P(C \cap B)\\
P\bigg\{|X+Y|> b\bigg\} &\leq P\left\{|X+Y|> b, |X|> \frac b2\right\} + P\left\{|X+Y|> b, |Y|> \frac b2\right\},
\end{align}
that is, the equality in the OP's original claim needs to be replaced by $\leq$ (as the OP noted in a comment on @aksakal's answer).
A: Note first that $|x+y| \le |x|+|y|$. This means that
$$A = \{(x,y): |x + y|>b\} \subseteq \{x: |x|>b/2\} \cap \{y: |y|>b/2\} =\\
\{(x,y): |x|>b/2 \textbf{ and } |y|>b/2\} = B$$
For your answer, it suffices to prove: 
$$A = \{(x,y): |x + y|>b\} \subseteq \{x: |x|>b/2\} \cup \{y: |y|>b/2\} =\\
\{(x,y): |x|>b/2 \textbf{ or } |y|>b/2\} = C$$
Since $A \subseteq B$ and $B \subseteq C$,then  $A \subseteq C$.
A: Firstly, thanks for everyone who answered my question! Here is the kind of proof I was asking for.
Let $A=\{(x,y):\mid x+y\mid \leq b\}$ and $B_1=\{(x,y):\mid x\mid \leq b/2\}$ and $B_2=\{(x,y):\mid y\mid \leq b/2\}$. Then
\begin{align}
B_1\cap B_2 &= \{(x,y):\mid x\mid \leq b/2,\mid y\mid \leq b/2\} \\
 &\subset \{(x,y):\mid x+y\mid \leq b,\mid y\mid \leq b/2\}\cup \{(x,y):\mid x+y\mid \leq b,\mid x\mid \leq b/2\}\\
 & \subset A.
\end{align}
Therefore 
\begin{equation}
A^c\subset B_1^c \cup B_2^c\\
\end{equation}
Now denote the preimages $\bar{A}=\{w:(X(w),Y(w))\in A^c\}$, $\bar{B}_1=\{w:(X(w),Y(w))\in B_1^c\}$ and $\bar{B}_2=\{w:(X(w),Y(w))\in B_2^c\}$, and note that 
$$w\in \bar{A} \iff (X(w),Y(w)) \in A^c \implies(X(w),Y(w)) \in B_1^c  \textrm{ 
 or } (X(w),Y(w)) \in B_2^c$$
using the previous inclusion. It means that $\bar{A}\subseteq \bar{B}_1\cup \bar{B}_2$. Only now we are in position to use the monotonicity of the measure
$$P(\bar{A})\leq P(\bar{B}_1\cup \bar{B}_2)\leq P(\bar{B}_1)+
 P(\bar{B}_2),$$
as desired.
