Define the notation $A(X) = \{ \alpha\colon P(|X|>\alpha)>0 \}$, so that $\|X\|_\infty = \sup A(X)$; we want to show that $\sup A(X+Y) \le \sup A(X) + \sup A(Y)$. Usually when trying to bound a sup from above, it's a good idea to name a generic element $\beta \in A(X+Y)$ and show that $\beta \le \sup A(X) + \sup A(Y)$. It suffices to show that there exists $c\in A(X)$ and $d\in A(Y)$ such that $\beta\le c+d$. (In some situations, it helps to show that for every $\epsilon>0$, one can find $c\in A(X)$ and $d\in A(Y)$ such that $\beta.)
So let $\beta\in A(X+Y)$, so that $P(|X+Y|>\beta)>0$. By the triangle inequality this implies that $P(|X|+|Y|>\beta)>0$. But notice that $|X|+|Y|>\beta$ if and only if there exist real numbers $c$ and $d$ such that $c+d>\beta$ and $|X|>c$ and $|Y|>d$; in fact, one can restrict $c$ and $d$ to be rational numbers. In other words, the event that $|X|+|Y|>\beta$ is contained in the union of the countable set of events $\big( |X|>c \text{ and } |Y|>d \big)\colon c,d\in\mathbb Q$. Since this countable union has positive probability, one of the individual events $\big( |X|>c \text{ and } |Y|>d \big)$ must itself have positive probability; in particular, each of $|X|>c$ and $|Y|>d$ has positive probability, so that $c\in A(X)$ and $d\in A(Y)$. Since $\beta, the proof is complete.