intuition on how to apply union bound For two events $A$ and $B$, then how to prove that

$$\Pr(A+B\geq 0) \leq \Pr(A\geq 0) + \Pr(B\geq 0)$$


I have a proof as follows, but I think it is not quite intuitive which needs to write it down to figure it out. 
\begin{align}
\Pr(A+B\geq 0) &= 1-\Pr(A+B < 0)\\\\
&\leq 1-\Pr(A<0 \cap B < 0)\\\\
& = \Pr(A\geq 0 \cup B \geq 0)\\\\
& \leq \Pr(A\geq 0) + \Pr(B\geq 0)
\end{align}
The second inequality by the observation that $Pr(A+B < 0) \geq \Pr(A<0 \cap B < 0)$, and the last step by union bound.
But this can not provide an intuition for me, is there simper proof or way to figure it out?
 A: As Zen says, you need to talk of random variables and not events.
$\{X+Y \geq 0\}$ is the event that the random point $(X,Y)$ lies on or above the diagonal line $x+y=0$ through the origin, which divides the plane into to two half-planes:the upper half-plane $\{(x,y)\colon x+y \geq 0\}$ and the lower half-plane
$\{(x,y)\colon x+y < 0\}$. Draw a sketch of the plane with coordinate axes $x$ and $y$ and mark on it the line $x+y=0$ to see why the half-planes are called so. In fact, mark the line $x=y$ on your sketch too and note that you have divided the plane into eight octants (numbered $1$ through $8$) and that the region of interest (the upper half-plane) is the disjoint union of the first, second, third, and eighth octants. Let $p_i$, $1 \leq i \leq 8$, denote the probability that $(X,Y)$ lies in the $i$-th octant.  Then, we have that
$$P\{X+Y \geq 0\} = p_1+p_2+p_3+p_8\tag{1}$$  Now, the half-plane $\{(x,y)\colon x \geq 0\}$ is the disjoint union of the first, second, seventh and eighth octants, while the half-plane  $\{(x,y)\colon y \geq 0\}$ is the disjoint union of the first through the fourth octants. Hence,
\begin{align} P\{X \geq 0\} &= p_1+p_2+p_7+p_8,\tag{2}\\
P\{Y \geq 0\} &= p_1+p_2+p_3+p_4.\tag{3}
\end{align}
Consequently,
\begin{align} P\{X + Y\geq 0\} &= p_1+p_2+p_3+p_8,\\
&\leq  (p_1+p_2+p_3)+p_4 + (p_8) + p_1+p_2+p_7\\
&= (p_1+p_2+p_3+p_4) + (p_1+p_2+p_7+p_8)\\
&= P\{Y \geq 0\} + P\{X \geq 0\}\\
\text{that is,} ~~P\{X + Y\geq 0\} &\leq P\{X \geq 0\} + P\{Y \geq 0\}.
\tag{4}
\end{align}
Furthermore, equality can hold in $(4)$ if and only if 
$p_1=p_2=p_4=p_7 = 0$. 
A: You need to reformulate this in terms of random variables $X$ and $Y$.
The reasoning is that if $X+Y\geq 0$, then either $X\geq 0$ or $Y\geq 0$; it can't be that both $X$ and $Y$ are negative. Hence, the event
$$
  \{X+Y\geq 0\}
$$
is a subset of
$$
  \{X\geq 0\}\cup\{Y\geq 0\}.
$$
Since $\Pr(\;\cdot\;)$ is a monotonic set function, we have
$$
  \Pr\{X+Y\geq 0\} \leq \Pr\left(\{X\geq 0\}\cup\{Y\geq 0\}\right)\leq\Pr\{X\geq 0\}+\Pr\{Y\geq 0\},
$$
in which the last inequality follows from the subadditivity of $\Pr(\;\cdot\;)$.
