Prove $|x - y|\le|x| + |y|$ (Spivak's Calculus Book)

Your approach is correct; do make sure that every step you have taken can also be taken in reverse. So far you have a proof that $$\textbf{If}\quad|x-y|\leq|x|+|y|\quad\textbf{then}\quad-xy\leq|xy|,$$ but of course you want to prove the opposite implication.

A shorter proof would be to use the triangle inequality; $$|x-y|=|x+(-y)|\leq |x|+|-y|=|x|+|y|.$$

Your proof is mostly correct, but you should mention that, for $a\ge0$ and $b\ge0$,

$a\le b$ if and only if $a^2\le b^2$

and that you can apply this because $|x-y|\ge0$ and $|x|+|y|\ge0$. Also you should say that the inequalities you write are equivalent to each other.

In the following steps, each line is equivalent to the one below it; also we use $|a|^2=a^2$ and that adding equal terms to both sides of an inequality doesn't change it. \begin{align} & |x-y|\le|x|+|y| \\[4px] & |x-y|^2\le(|x|+|y|)^2 \\[4px] & x^2-2xy+y^2\le x^2+2|x|\,|y|+y^2 \\[4px] & {-}2xy\le |2xy| \end{align} Since the last inequality is true (because $-a\le|a|$ for every $a$), we end the proof.