Let $f$: $(E,d) \to (E',d')$. Assume that $d'(f(x),f(y)) \leq (d(x,y))^2$ $\forall\ x,y \in E$. Assume $E$ is bounded. Prove that $f$ is continuous.
This is what I have come up with so far but I'm not sure it is $100\%$ correct/a good use of the fact the metric space is bounded.
Proof: Let $\varepsilon > 0$. Since $E$ is bounded, $a = g.l.b(E)$ and $b = l.u.b.(E)$ s. t. $a < x < b$ and $a < y < b$ $\forall x,y \in E$. Thus, since $E$ is bounded $\exists\ d(x,y) = |x - y| < \varepsilon$ $\implies d(x,y)^2\ <\ \varepsilon^2$. We choose $\delta\ =\ \varepsilon^2$. Then it follows that $d'(f(x), f(y)) \leq (d(x,y))^2 < \varepsilon^2 = \delta$. Thus, $f(x)$ is uniformly continuous.
Any tips? How do I make this more concrete/did I bark up the wrong tree?
$E$ is bounded here means that, for every $y\in E$, $M_{y}:=\sup_{x\in E}d(x,y)<\infty$. Now $d'(f(x),f(y))\leq M_{y}d(x,y)$, so $f$ is continuous at $y$.