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?

Hint: verify
$$ \{ |X+Y|\geq \varepsilon\} \subset \{ |X|\geq \varepsilon/2 \}\cup \{ |Y|\geq \varepsilon/2 \}$$
and use finite subadditivity.
