We first show that the sequence is uniformly bounded. By uniform equicontinuity, there is a \(\delta > 0\) such that for all \(x \in X\) and all \(n \in \N\text{,}\)
\begin{equation*}
B(x,\delta) \subset f_n^{-1}\bigl(B(f_n(x),1)\bigr) .
\end{equation*}
The space \(X\) is compact, so there exist \(x_1,x_2,\ldots,x_k\) such that
\begin{equation*}
X = \bigcup_{j=1}^k B(x_j,\delta) .
\end{equation*}
As \(\{ f_n \}_{n=1}^\infty\) is pointwise bounded there exist \(M_1,M_2,\ldots,M_k\) such that for \(j=1,2,\ldots,k\text{,}\)
\begin{equation*}
\sabs{f_n(x_j)} \leq M_j \qquad \text{for all } n.
\end{equation*}
Let \(M \coloneqq 1+ \max \{ M_1,M_2,\ldots,M_k \}\text{.}\) Given any \(x \in X\text{,}\) there is a \(j\) such that \(x \in B(x_j,\delta)\text{.}\) Therefore, for all \(n\text{,}\) we have \(x \in f_n^{-1}\bigl(B(f_n(x_j),1)\bigr)\text{,}\) or in other words
\begin{equation*}
\sabs{f_n(x)-f_n(x_j)} < 1 .
\end{equation*}
By the reverse triangle inequality,
\begin{equation*}
\sabs{f_n(x)} < 1+ \sabs{f_n(x_j)} \leq 1+M_j \leq M .
\end{equation*}
As \(x\) was arbitrary, \(\{f_n\}_{n=1}^\infty\) is uniformly bounded.
Next, pick a countable dense subset
\(D \subset X\text{.}\) By
Proposition 11.6.5, we find a subsequence
\(\{ f_{n_j} \}_{j=1}^\infty\) that converges pointwise on
\(D\text{.}\) Write
\(g_j \coloneqq f_{n_j}\) for simplicity. The sequence
\(\{ g_n \}_{n=1}^\infty\) is uniformly equicontinuous. Let
\(\epsilon > 0\) be given, then there exists a
\(\delta > 0\) such that for all
\(x \in X\) and all
\(n \in \N\text{,}\)
\begin{equation*}
B(x,\delta) \subset g_n^{-1}\bigl(B(g_n(x),\nicefrac{\epsilon}{3})\bigr).
\end{equation*}
By density of \(D\) and because \(\delta\) is fixed, every \(x \in X\) is in \(B(y,\delta)\) for some \(y \in D\text{.}\) By compactness of \(X\text{,}\) there is a finite subset \(\{ x_1,x_2,\ldots,x_k \} \subset D\) such that
\begin{equation*}
X = \bigcup_{j=1}^k B(x_j,\delta) .
\end{equation*}
As \(\{ x_1,x_2,\ldots,x_k \}\) is a finite set and \(\{ g_n \}_{n=1}^\infty\) converges pointwise on \(D\text{,}\) there exists a single \(N\) such that for all \(n,m \geq N\text{,}\)
\begin{equation*}
\sabs{g_n(x_j)-g_m(x_j)} < \nicefrac{\epsilon}{3}
\qquad \text{for all } j=1,2,\ldots,k.
\end{equation*}
Let \(x \in X\) be arbitrary. There is some \(j\) such that \(x \in B(x_j,\delta)\) and so for all \(\ell \in \N\text{,}\)
\begin{equation*}
\sabs{g_\ell(x)-g_\ell(x_j)} < \nicefrac{\epsilon}{3}.
\end{equation*}
So for \(n,m \geq N\text{,}\)
\begin{equation*}
\begin{split}
\sabs{g_n(x)-g_m(x)} & \leq
\sabs{g_n(x)-g_n(x_j)} +
\sabs{g_n(x_j)-g_m(x_j)} +
\sabs{g_m(x_j)-g_m(x)}
\\
& <
\nicefrac{\epsilon}{3} +
\nicefrac{\epsilon}{3} +
\nicefrac{\epsilon}{3} = \epsilon .
\end{split}
\end{equation*}
Hence, \(\{ g_n \}_{n=1}^\infty\) is uniformly Cauchy. By completeness of \(\C\text{,}\) it is uniformly convergent.