PROOF: Assume $x$ and $y$ are real numbers. Consider the following cases.
CASE 1: assume $x \ge 0$ and $y \ge 0$
CASE 2: assume $x \ge 0$ and $y < 0$
CASE 3: assume $x < 0$ and $y \ge 0$
CASE 4: assume $x < 0$ and $y < 0$
What steps should be taken from here? I assume my cases are correct, I am just having trouble fleshing them out.
As you suggested, we can examine case by case. If both $x$ and $y$ are positive, or both $x$ and $y$ are negative, then the statement is true. (Simply replace $x$ with $-x$ or $y$ with $-y$, and compute.) Now consider the case whether one of $x$ and $y$ is negative. WLOG, let $y = -n$. Then, we have $|x-n| \leq x + n$. If $|x-n|$ is positive, then $x -n$ is less than $x+n$, which is clearly true. Similarly, if $|x-n|$ is negative, then we have $n - x$ is less than $x + n$, which implies $x > 0$, which is true. Therefore, we are done.
Your cases are exhaustive, but since replacing $x,\,y$ with $-x,\,-y$ doesn't change $|x+y|$ or $|x|+|y|$, you actually only need to check case 1 (which then implies case 4 works) and case 2 (which then implies case 3 works). For case 1, $x+y\ge0$ and the modulus signs can be dropped, so the inequality is true (in fact, it reduces to an equation). For case $2$, we subdivide into the options $x\ge |y|,\,x<|y|$. The first gives $|x+y|=|x|-|y|\le|x|+|y|$, the second $|x+y|=|y|-|x|\le|x|+|y|$.
