I'm having trouble finishing this. One approach that I made is this:
Let $\epsilon > 0$. Then, since $f$ is continuous, for every $x \in K$ exists $\delta_x > 0$ such that $d(x, x')<\delta_x \Rightarrow d'(f(x),f(x')) < \epsilon$.
Then $B=\{B(x, \delta_x)\}_{x\in K}$ covers $K$. Since $K$ is compact, I can take finite amount of this $B_x$ and they cover $K$. Let's say that $K = \cup_{i=1}^n B(x_i, \delta_{x_i})$. Let $\delta = min\{\delta_{x_i}\}$ I would like here to say that if $d(x, y) < \delta \Rightarrow d'(f(x), f(y)) < \epsilon$. But I can't. If $x,y \in B(x_i, \delta_{x_i})$ for some $i$, then everything is fine. But I can take two point $x,y$ as close as I want, but being on different open sets.
Best Answer
You are close. Try this modification: fix $\varepsilon>0$. We can take an open cover $B_{\delta_x}(x)$ around every $x\in K$ so that $d(x,x')<\delta_x$ implies $d'(f(x),f(x'))<\varepsilon$. Now, we can also take an analogous open cover $\{B_{\delta_x/2}(x):x\in X\}$. Note that there exists a finite set of indices $i=1,\ldots, n$ such that the $B_{\delta_{x_i}/2}(x_i)$ cover $K$ by compactness. A fortiori, the same holds for the $B_{\delta_{x_i}}(x_i)$. Define $$\delta=\min\{\delta_{x_i}/2: i=1,\ldots,n\}.$$ Then if $x,y\in K$ are given so that $d(x,y)<\delta$, note that $x\in B_{\delta_{x_i}/2}(x_i)$ for some $i$. Now, $$ d(y,x_i)\le d(y,x)+d(x,x_i)<\delta+\frac{\delta_{x_i}}{2}\le\frac{\delta_{x_i}}{2}+\frac{\delta_{x_i}}{2}<\delta_{x_i}$$ so that $y\in B_{\delta_{x_i}}(x_i)$ as is $x$. Then $d'(f(x),f(x'))<\varepsilon$ by the beginning part.