See: http://artofproblemsolving.com/community/c7h396247p2203339

$x \to 1, y \to 0$ : $|f(1)-f(0)|=1$. When $f$ is a solution, $g=\pm f + C$ is also a solution. So, it is enough to see when $f(0)=0, f(1)=1$. $x \to 0$ : $|f(y)| \leq y$, $x \to 1$: $|1-f(y)| \leq 1-y$. Adding two inequality, $1 \leq |f(y)|+|1-f(y)| \leq 1$. Thus $f(y)=y$ and the general solution is $f(x)=\pm x + C$.