To prove $$\sup\left\{\alpha\mid P(|X+Y|>\alpha)>0 \right\}\leq||X||_\infty+||Y||_\infty$$ it suffices to prove for every $\alpha$ that
$$P(|X+Y|>\alpha)>0 \Longrightarrow ||X||_\infty+||Y||_\infty \ge\alpha$$
and for that we only need to prove for arbitrary $\epsilon>0$ that
$$P(|X+Y|>\alpha)>0 \Longrightarrow ||X||_\infty+||Y||_\infty\ge\alpha-4\epsilon$$
Thus, assume the left-hand side, and consider the conditional probablity distribution of $(|X|,|Y|)\in\mathbb R^2$ given $|X+Y|>\alpha$. Since $\mathbb R^2$ is covered by countably many balls of radius $\epsilon$, there must be one such ball that contains positive probability mass, so there exist $x,y$ such that
$$P(|X|>x-\epsilon)>0 \land P(|Y|>y-\epsilon)>0$$
and hence $||X||_\infty+||Y||_\infty\ge x+y-2\epsilon$.
Furthermore, we must have $x+y\ge\alpha-2\epsilon$, because otherwise the triangle inequality prevents the ball from containing any $(|X|,|Y|)$ with $|X+Y|>\alpha$. Therefore
$$||X||_\infty+||Y||_\infty\ge\alpha-4\epsilon$$
as promised.