Prove formally that $P(|X+Y|>\varepsilon)\leq P(|X|>\frac{\varepsilon}{2})+ P(|Y|>\frac{\varepsilon}{2})$ I'm stuck with a simple probability inequality. For arbitrary random variables 
$$P(|X+Y|>\varepsilon)\leq P(|X|>\frac{\varepsilon}{2})+ P(|Y|>\frac{\varepsilon}{2}) $$
Using $P(A)+P(B) \geq P(A \cap B) = P(X,Y > \frac{\varepsilon}{2})$ I get underestimate, as $\{|X+Y|>\varepsilon\}$ is satisfied not only for such values of $X,Y$, so it should be bigger. The probability $P(A)+P(B) \geq P(A \cup B)=P(X \text{ or }Y > \frac{\varepsilon}{2})$ looks like what's needed as it is the bigger event and can contain the values that will add up to the value $<\varepsilon$. But how to pass from heuristics to the formal proof?

First, observe that $$\{|X+Y|>\varepsilon\}\subseteq\left\{|X|>\frac{\varepsilon}{2}\right\}\cup\left\{|Y|>\frac{\varepsilon}{2}\right\}.\tag{$\star$}$$ Indeed, if $|X+Y|>\varepsilon$, then either $|X|>\varepsilon/2$ or $|Y|>\varepsilon/2$ must hold, for if both $|X|\leq\varepsilon/2$ and $|Y|\leq\varepsilon/2$ held, then $|X+Y|\leq|X|+|Y|\leq\varepsilon/2+\varepsilon/2=\varepsilon$ would also hold by virtue of the triangle inequality.
Now, ($\star$) implies, by monotonicity and subadditivity of probability measures, that
\begin{align*}
\mathbb P(\{|X+Y|>\varepsilon\})\leq&\,\mathbb P\left(\left\{|X|>\frac{\varepsilon}{2}\right\}\cup\left\{|Y|>\frac{\varepsilon}2\right\}\right)\\
\leq&\,\mathbb P\left(\left\{|X|>\frac{\varepsilon}{2}\right\}\right)+\mathbb P\left(\left\{|Y|>\frac{\varepsilon}2\right\}\right).
\end{align*}
