Proof for Uniform Convergence for $\{f_n\}$ 
Suppose $\{f_n\}$ is an equicontinuous sequence of functions defined on $[0,1]$ and $\{f_n(r)\}$ converges $∀r ∈ \mathbb{Q} \cap [0, 1]$. Prove that $\{f_n\}$ converges uniformly on 
  $[0, 1]$.

Since I know that $\mathbb{Q} \cap [0, 1]$ is not compact, I am a bit stuck on my proof.
So far I have:
Let $f_n \to f$ pointwise on $\mathbb{Q} \cap [0, 1]$ 
Since $\{f_n\}_n$ is equicontinuous and point-wise bounded (it’s pointwise convergent, so in particular), there exists a subsequence $\{f_{n_k}\}_k$ such that $f_{n_k} \to f$ uniformly. 
Since each $f_n$ is continuous, $f$ is then continuous.
Now take  $\varepsilon > 0$. Using equicontinuity of $\{f_n\}_n$, we find $\delta_1 > 0$ such that if $d(x, y) < δ_1$, $x, y \in K$, then
$|f_n(x) − f_n(y)| < \varepsilon/3$ for all $n ∈ \mathbb{Z}^+$. 
Using continuity of $f$, for each $x \in K$, let $\delta_2 = \delta_2(x) > 0$ be such that if $|x − y| < \delta_2(x)$, $y \in \mathbb{Q} \cap [0, 1]$, then $|f(x) − f(y)| < \varepsilon/3$. For $x \in \mathbb{Q} \cap [0, 1]$, let $\delta(x) = \min(\delta_1, \delta_2(x)) > 0$
I am not sure how to continue nor am I too sure I am on the right path.

This is actually a simple case of applying the Arzela-Ascoli Propagation Theorem which states:
Point-wise convergence of an equicontinuous sequence of functions on a dense subset of the domain propagates to uniform
convergence on the whole domain.
The rational numbers, $\mathbb{Q}$, are dense in the interval $[0,1]\subset \mathbb{R}$. So if <${f_n(r)}$> is converging to some function $f$ for every $r\in\mathbb{Q}$ $\cap$ $[0,1]$, the convergence will propagate to uniform convergence on all of $[0,1]$.
A proof of this theorem can be found on page 227 of Real Mathematical Analysis by Charles Pugh.
