Suppose that $M$ is covering compact and that $f: M \to N$ is continuous. Use the Lebesgue number lemma to prove that $f$ is uniformly continuous.
Given $\epsilon >0$, the set $\mathcal{U} = \{B_{\epsilon/2}(p) \mid p \in N\}$ is an open covering of $f(M)$. Since $f$ is continuous, $f^{-1}(B_{\epsilon/2}(p))$ is open. But $M$ is compact, then $f(M)$ too. Thus, we can find a finite subcovering
$$f(M) \subset B_{\epsilon/2}(p_{1})\cup \cdots \cup B_{\epsilon/2}(p_{n}).$$
Therefore,
$$f^{-1}(B_{\epsilon/2}(p_{1}))\cup \cdots \cup f^{-1}(B_{\epsilon/2}(p_{n}))$$
is a finite subcovering of $M$. By the Lebesgue nummber lema, there is a $\delta>0$ such that for every $x \in B_{\delta}(x) \subset f^{-1}(B_{\epsilon/2}(p_{i}))$ for some $i$. So, if $d(x,y) < \delta$, the "$y$" are in some $B_{\delta}(x) \subset f^{-1}(B_{\epsilon/2}(p_{i}))$, then $d(f(x),f(y)) \leq d(f(x),p) + d(f(y),p) < \epsilon.$
Is this correct?
Best Answer
The Lebesgue number lemma I know (e.g. as stated here) needs no finite cover but any open cover.
So just use the open cover $\{f^{-1}[B(f(x), \frac{\varepsilon}{2})]: x \in M\}$ for $M$ and find $\delta >0$ such that any set with diameter $< \delta$ sits inside one member of the cover.
If then $d(x,x') < \delta$ apply this to $\{x,x'\}$, so that for some $p \in M$ we have $\{x,x'\} \subseteq f^{-1}[B(f(p), \frac{\varepsilon}{2})]$ which means $f(x), f(x') \in B(f(p), \frac{\varepsilon}{2})$ and then the triangle inequality via $f(p)$ yields $d(f(x), f(x')) < \varepsilon$, as required.