Baby Rudin Problem 7.16 

Problem 7.16: Suppose $\{f_n\}$ is an equicontinuous family of functions on a compact set $K$ and $\{f_n\}$ converges pointwise to some $f$ on $K$. Prove that $f_n \to f$ uniformly.


Now for this problem I assume that $f_n,f : K \subset \Bbb{R} \rightarrow \Bbb{R}$ with the usual euclidean metric. Even though this is not assumed in the problem, I assume this to simplify matters first.
Now I believe I have proven that $f_n$ is uniformly cauchy on $K$ as follows. By equicontinuity of the family $\{f_n\}$ I can choose $\delta> 0$ such that $|x-p_i|< \delta$ will imply that $|f_m(x) - f_m(p_i)| < \epsilon$, and $|f_n(x) - f_n(p_i)| < \epsilon$.
Consider the collection $\{B_\delta(x)\}_{x \in K}$ that clearly covers $K$, by compactness of $K$ we get that there are finitely many points $p_1,\ldots p_n \in K$ such that $\{B_\delta(p_i)\}_{i=1}^n$ is a cover for $K$. Furthermore, because $f_n \rightarrow f$ pointwise for each $x \in K$ we get a cauchy sequence of numbers, so in particular given any $\epsilon > 0$, for each $p_i$ there exists $N_i$ such that $m,n \geq N_i$ implies that $|f_m(p_i) - f_n(p_i) | < \epsilon$. Taking 
$$N = \max_{1 \leq i \leq n} N_i$$
gives that $m,n\geq N$ implies that $|f_m(p_i) - f_n(p_i)| < \epsilon$ for all $i$. 
Now we can finally put everything together to prove uniform cauchyness, take any $x \in K$ so that $x \in B_\delta(p_j)$ for some $1 \leq j \leq n$. Then
$$\begin{eqnarray*} |f_n(x) - f_m(x)| &\leq& |f_n(x) - f_n(p_i) | + | f_n(p_i) - f_m(p_i)| + |f_m(p_i) - f_m(x)| \\
&<& \epsilon + \epsilon + \epsilon \\
&=& 3\epsilon. \end{eqnarray*}$$
The first and last term being less than $\epsilon$ come from equicontinuity, the middle term being less than $\epsilon$ comes from the derivation just before. Now what I am thinking of doing now to prove uniform cauchyness is to take the sup on the left, is this something legal I can do? Also are there are any mistakes in the proof above?

Here is some context why I want to prove uniform cauchyness: Suppose I know that $\{f_n\}$ is uniformly cauchy. Then I know that given any $\epsilon > 0$, there exists $N \in \Bbb{N}$ such that $m,n\geq N$ implies that $|f_n(x) - f_m(x)| < \epsilon$ for all $x \in K$. Now we do this trick of fixing one of the indices. Fix $n$ to be some integer greater than $N$ and let $m\rightarrow \infty$, we see that
$$\begin{eqnarray*} |f_n - f| &=& \lim_{m\rightarrow \infty} | f_n - f_m| \\
&\leq& \epsilon \end{eqnarray*} $$
by the limit comparison test. Recall that $f$ was the pointwise limit of $\{f_n\}$. But then since $n$ was any arbitrary integer greater than $N$ we have that $f_n \rightarrow f$ uniformly.
 A: Yes this looks ok. The point with the  $3\varepsilon$ is, that the rhs does not depend on $x$, and then the lhs is estimated for all $x$ and sufficiently large $n,m$ simultaneously. That's the important thing to see here. 
You could, alternatively, try to work with $f$ and $f_n$ right from the beginning -- the Cauchy sequence approach is correct, but I'd consider it a detour. Depends on your liking alone, though.
A: Your argument is great until the point where you say "Now what I am thinking of doing..."
You have already proven that $\{f_n\}$ is uniformly Cauchy since you've shown that for any $\epsilon>0$, there is an $N$ so that if $n,m>N$, $|f_m(x)-f_n(x)|<3\epsilon$ for all $x\in K$.
To show uniform convergence, we simply need to say that, because of the pointwise convergence, for any $x$, there is some $M>N$ so that if $m>M$, $|f_m(x)-f(x)|<\epsilon$. Then, for any $n>N$, we have that
$$
\begin{align}
|f_n(x)-f(x)|
&\le|f_n(x)-f_m(x)|+|f_m(x)-f(x)|\\
&<3\epsilon+\epsilon\\
&=4\epsilon
\end{align}
$$
That is, for any $\epsilon>0$, there is an $N$ so that if $n>N$, $|f_n(x)-f(x)|<4\epsilon$ for all $x\in K$.
