Let $f: \Bbb R^2 \to \Bbb R$ be a continuous function such that $f(x,0) = 0,$ for all $x$. Show that exists $r > 0$ such that $|f(x,y)| \leq 1/4$ whenever $(x,y) \in [0,1] \times [-r,r]$.
Attempt: If $f$ is uniformly continuous, it is pretty easy.. let $\epsilon = 1/4 > 0$. So exists $r > 0$ such that: $$\|(x,y) - (x,0)\| \leq r \implies |f(x,y)-f(x,0)| \leq 1/4,$$ but that reads $|y| < r \implies |f(x,y)| \leq 1/4$, and we're happy. If I only assume that $f$ is continuous, maybe I could restrict $f$'s domain to a compact set. So $f$ would be uniformly continuous in this compact, and by the argument above I'm done. I do know that for every $r > 0$, $[0,1] \times [-r,r]$ is compact. But I want to prove that such $r$ exists. If $f$ is only continuous, that $r$ is varying, and the supremum of all of these $r$'s may not exist. Can someone please give me some pointers? Thanks.
Hint : $f$ is not uniformly continuous on ${\mathbb R}^2$ in general, but it is uniformly continuous on compact sets such as $[0,1] \times [-1,1]$.