Let $A \subset C^0([a,b],\mathbb{R})$
Show that if $A$ is equicontinuous, then $\overline A$ is equicontinuous.
Preliminary Proof:
Let $(f_n) \subset \overline A$, let $\epsilon >0$ be given, and $|x-y| < \delta$
$|f_n(x) - f_n(y)| \leq |f_n(y) - g_k(y)| + |g_k(y) - g_k(x)| + |g_k(x) - f_n(x)|$
where $(g_k)$ is an equicontinuous sequence of continuous functions in $A$ that approaches $f_n$ by definition of closure.
Recall, $g_k \rightrightarrows f_n$ if $\forall \epsilon > 0, \exists K_x$ s.t. $\forall x \in [a,b], \forall k \geq K_x, |g_k(x) - f_n(x)| < \epsilon$
Take $k = \max\{K_x, K_y\}$, thus $|f_n(y) - g_k(y)|< \dfrac{\epsilon}{3}, |g_k(x) - f_n(x)| < \dfrac{\epsilon}{3}$ and by equicontinuity $|g_k(y) - g_k(x)|< \dfrac{\epsilon}{3}$
Then $|f_n(x) - f_n(y)| < \dfrac{\epsilon}{3} + \dfrac{\epsilon}{3} + \dfrac{\epsilon}{3} < \epsilon, \forall n$. Thus $\overline A$ is equicontinuous.
Question:
Since $A$ is just a subset of $C^0$, not necessarily a sequence, how to modify the proof so we can drop dependence on $n$ and $k$
Please let me know if the proofs have any other serious flaws
As you have sensed, the first step, where you let $f_n$ be a sequence in $\bar A,$ is off the mark. (Note also the notation $(f_n) \subset \bar A$ is not quite right.)
You want to show that the $\delta$ that works on $A$ for a given $\epsilon$ will also work on $\bar A.$ That's not quite right, but that's got to be the intuition.
So let $\epsilon>0.$ Then there exists $\delta > 0$ such that $|x-y|<\delta$ implies $|f(x)-f(y)|<\epsilon$ for all $f\in A.$ Now really, how far off could any $f$ that is the uniform limit of a sequence in $A$ be from satisfying this?