Let $K$ be compact, if $\{f_n\}$ is point wise bounded and equicontinuous on $K$, then $\{f_n\}$ contains a uniformly convergent subsequence.
Question:
I understand the idea (if compact in $\mathbb{R}^n$) of constructing a subsequence such that it converges whenever $x$ is a rational number. Then extend to real, so in order to prove uniform, at least we have point wise.
However, my question is, in general, why do we need a dense subset of the compact domain? If compact, then with (uniformly) equicontinuous we have a $\delta$ net. By B-W theorem, we can construct a convergent subsequence that converges at those centers of such a $\delta$ net. How could I run into trouble from here? (without involving dense subset.)
Sure, you can do that. To flesh it out, given $\epsilon > 0$, choose a $\delta$ from the assumption of equicontinuity. Take a $\delta$ net and find a subsequence $f_{n_k}$ that converges at each point of your net. Conclude that $\limsup_{j,k \to \infty} |f_{n_j} - f_{n_k}| < \epsilon$.
But the subsequence you produced depended on $\epsilon$, and you want a single subsequence that works for all $\epsilon$. So you need a "diagonal subsequence" or Tychonoff type argument now. That will give you a single subsequence $f_{m_k}$ for which $\limsup_{j,k \to \infty} |f_{m_j} - f_{m_k}| \to 0$, i.e. the subsequence is uniformly Cauchy. Now you have to invoke the uniform completeness of the space of continuous functions, and you are done.
It didn't really save a lot. You still had to use the same "diagonal subsequence" argument that gets used in proving that you can find a subsequence converging at every point of your countable dense subset, so that wasn't avoided.
And here you had to use the uniform completeness theorem to assert the existence of a limiting function. With the dense subset approach, you would avoid that. Once you have a subsequence $f_{n_k}$ converging at every rational, you have a fairly explicit definition of the limiting function $f$: it's the unique continuous function which is given at the rationals $q$ by $f(q) = \lim_{k \to \infty} f_{n_k}(q)$.
Note that you also didn't strengthen the result: every compact metric space has a countable dense subset (proof: for each $n$ take the points of a $1/n$ net), so the existence of such a subset isn't an extra assumption.