Let $K$ be a compact subset of $\mathbb{R}^n$. Let $\{f_n\}$ be a sequence of contractions on $K$. We need to show that $\{f_n\}$ has a uniformly convergent subsequence.
Initially, I struggled to understand what we have to show. Here I will lay out what we know first. That a sequence of contractions $\{f_n\}$ has a uniformly convergent subsequence must mean that the set $C(K,K)$ is compact. Observe that $C(K,K) = \{f: K \to K \,|\, f \text{ is bounded and continuous} \}$. By the Arzelà–Ascoli theorem, this means that $K \subseteq C$ is closed, bounded, and equicontinuous. The solution hint suggests choosing $K'$ to be the closure of $C(K,K)$ (Why is this? Is it because it makes it easier to define the set for which we are trying to prove compactness?)
Accepting this fact, the objective then is to prove that $K'$ is compact:
Closure: Of course, every closure of a set is closed.
Boundedness: Since $K$ is compact, it must be bounded. This means that there exists an $L \in \mathbb{R}$ such that for all $x, y \in K$, $|x-y| \leq L$. As any two functions in $K'$ map from $K$ to $K$, $\sup_{x \in K} |f(x) - g(x)| \leq \sup_{x, y \in K} |x-y| \leq L$. This is true for any $f, g \in K'$, so $K'$ is bounded.
Equicontinuity: For all $f \in K'$, $|f_n(x) - f_n(y)| < M_n |x-y|$ for all $n \in \mathbb{N}$. Then $|f(x) - f(y)| = \lim_{n \to \infty} |f_n(x) - f_n(y)| \leq \sup_{n \in \mathbb{N}} M_n |x-y| \leq |x-y|$. So, given $\epsilon > 0$, we let $\delta = \epsilon$, and we get what we sought. Therefore, $K'$ is compact and we are done.
Any corrections on this proof, in addition to answering the question I posed above, namely, why do we choose $K'$ to be the closure to proceed with the proof?