Let $(S,d)$ be a metric space. Assume that $f:S \rightarrow \mathbb R$ and $g:S \rightarrow \mathbb R$ are uniformly continuous on $S$. Prove that $h:S \rightarrow \mathbb R^2$ defined by $h(s)=(f(s),g(s))$ is uniformly continuous on $S$ where $\mathbb R^2$ has the Euclidean metric.
I know by definition, that for metric spaces $(S,d), (S^*,d^*)$ if $f:S \rightarrow S^*$ is uniformly continuous on $S$, then for each $\epsilon > 0$, there exists a $\delta>0$ such that $s,t \in S$ and $d(s,t)<\delta$ imply that $d^*(f(s),f(t))<\epsilon$
So, that means that for $f,g$, we have this property. Thus, given $\epsilon$ suppose we have $s,t \in S$ and $d(s,t)<\delta$, then $d^*((f(s),g(s)), (f(t),g(t)))=\sqrt {(f(s)-f(t))^2-(g(s)-g(t))^2}$.
I think what I have so far is correct, but I'm not sure how to proceed.
$=\sqrt {(f(s)-f(t))^2+(g(s)-g(t))^2}< \sqrt {\epsilon^2+\epsilon^2}=\sqrt{2}\epsilon.$