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.
 A: Step 1: For every $x\in [0,1]$, $\{f_n(x)\}_n$ converges. 
Fix $\varepsilon >0$. Pick $\delta>0$ witnessing the definition of equicontinuity for $\varepsilon/3$. Pick a rational number $r$ with $|x-r|<\delta$. Fix $N$ such that $|f_n(r)-f_m(r)|<\varepsilon/3$ for every $n,m\ge N$. 
If $n,m\ge N$, then 
$$
|f_n(x)-f_m(x)|\le |f_n(x)-f_n(r)|+|f_n(r)-f_m(r)|+|f_m(r)-f_m(x)|\le \varepsilon
$$
Thus $\{f_n(x)\}_n$ is a Cauchy sequence, and we're done by equicontinuity. 
Let $f(x):=\lim_n f_n(x)$. 
Step 2: The convergence is uniform. See this answer
A: Firstly, we show that $\lim_{n\rightarrow\infty}f_{n}(x)$ exists
for all $x\in[0,1]$. Let $x\in[0,1]$. Let $\varepsilon>0$ be arbitrary.
By equicontinuity, there exists $\delta>0$ such that $|f_{n}(x)-f_{n}(y)|<\varepsilon$
whenenver $n\in\mathbb{N}$ and $y\in[0,1]\cap(x-\delta,x+\delta)$.
By density of $\mathbb{Q}$, there exists $r\in[0,1]\cap(x-\delta,x+\delta)$.
Choose $N$ such that $|f_{n+p}(r)-f_{n}(r)|<\varepsilon$ whenever
$n\geq N$ and $p\in\mathbb{N}$ (this is possible because $\{f_{n}(r)\}_{n}$
is convergent). For any $n\geq N$ and $p\in\mathbb{N}$, we have
\begin{eqnarray*}
 &  & |f_{n+p}(x)-f_{n}(x)|\\
 & \leq & |f_{n+p}(x)-f_{n+p}(r)|+|f_{n+p}(r)-f_{n}(r)|+|f_{n}(r)-f_{n}(x)|\\
 & < & 3\varepsilon.
\end{eqnarray*}
This shows that $\{f_{n}(x)\}$ is a Cauchy sequence and hence it
converges.
Next, we show that $\{f_{n}(x)\}$ converges uniformly in $x$. Let
$\varepsilon>0$ be arbitrary. By equicontinuity, for each $x\in[0,1]$,
there exists $\delta_{x}>0$ such that $|f_{n}(x)-f_{n}(y)|<\varepsilon$
whenever $y\in[0,1]\cap(x-\delta_{x},x+\delta_{x})$ and $n\in\mathbb{N}$.
Note that $\{(x-\delta_{x},x+\delta_{x})\mid x\in[0,1]\}$ is an open
convering for the compact set $[0,1]$, so it has a finite subcover
$\{(x_{i}-\delta_{x_{i}},x_{i}+\delta_{x_{i}})\mid i=1,\ldots,K\}$.
Choose $N$ such that $|f_{n+p}(x_{i})-f_{n}(x_{i})|<\varepsilon$
whenever $n\geq N$, $p\in\mathbb{N}$, and $i=1,2,\ldots,K$. Now,
let $x\in[0,1]$, $n\geq N$, and $p\in\mathbb{N}$ be arbitrary. Choose $i$ such that $x\in (x_{i}-\delta_{x_{i}},x_{i}+\delta_{x_{i}})$.
We have 
\begin{eqnarray*}
 &  & |f_{n+p}(x)-f_{n}(x)|\\
 & \leq & |f_{n+p}(x)-f_{n+p}(x_{i})|+|f_{n+p}(x_{i})-f_{n}(x_{i})|+|f_{n}(x_{i})-f_{n}(x)|\\
 & < & 3\varepsilon.
\end{eqnarray*}
This shows that $\{f_{n}(x)\}$ is uniformly Cauchy in $x$ and hence $\{f_n\}$
 converges uniformly.
A: 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.
