I am interested in a proof of the following statement:
Let $(X,d_X)$ be a compact metric space and $(Y,d_Y)$ be a complete space. Then for $f,g\in C (X,Y)$ the metric $$d(f,g):=\sup_{x\in X}d_Y(f(x),g(x))$$ turns $(C(X,Y),d)$ into a complete metric space.
However, I can't find this statement in any of my (beginner's) books and also searching on the internet hasn't turned up anything. I would be thankful if someone could provide a script or notes where I could read it. Unfortunately, my mathematical abilites are far from trying to prove it on my own (I would be happy to just be able to understand it) and I hope that asking for a complete proof on this page isn't inappropriate.
Let $(f_n)_{n\in\mathbb N}$ be a Cauchy sequence of elements of your space. I will prove that it converges.
Let $x\in X$. For each $\varepsilon>0$, there is a natural $p$ such that$$m,n\geqslant p\implies d_Y\bigl(f_m(x),f_n((x)\bigr)\leqslant d(f_m,f_n)<\varepsilon.$$Therefore, $\bigl(f_n(x)\bigr)_{n\in\mathbb N}$ is a Cauchy sequence of elements of $Y$ and therefore it converges. Define$$f(x)=\lim_{n\to\infty}f_n(x).$$
Now, I will prove that $(f_n)_{n\in\mathbb N}$ converges to $f$. Take $\varepsilon>0$ and take $\varepsilon'\in(0,\varepsilon)$. There is some natural $p$ such that$$m,n\geqslant p\implies d(f_m,f_n)<\varepsilon'.$$So, if you fix $n\in\mathbb N$, then, for each $x\in X$,$$d_Y\bigl(f(x),f_n(x)\bigr)=\lim_{m\to\infty}d_Y\bigl(f(x),f_n(x)\bigr)\leqslant\varepsilon'.$$Therefore, $d(f,f_n)\leqslant\varepsilon'<\varepsilon$. What this means is, of course, that $\lim_{n\to\infty}f_n=f$.