Uniform convergence on $(a,b)\cap \mathbb{Q}\implies $ Uniform convergence on $[a,b]$ 
If $f_n$ is continuous on $[a,b]$ for each $n$ and it converges uniformly on $(a,b)\cap \mathbb{Q}$, prove it converges uniformly on $[a,b]$

I'm trying to solve this problem and from the given part of the problem. I felt that I should use Dini's theorem to get the result, but I'm not sure how $f_n$ converges uniformly on $(a,b) \cap \mathbb{Q}$ comes to use. Am I thinking it in the wrong approach?
 A: $$|f_n(x)-f_m(x)|\leq |f_n(x)-f_n(r)|+|f_n(r)-f_m(r)|+|f_m(r)-f_m(x)|$$
Ok, so given $x$, and $n,m\geq N$ chose $r$ rational so that the first and third terms are less than $\epsilon$. The middle term will be less than $\epsilon$ for all $r$, rational. Where of course $N$ is chosen large enough. 
A: Each  $f_n$ is continuous on $[a,b]$ and $f_n\to f$(say) uniformly on  $(a,b)\cap \Bbb Q$ so  $f$ is continuous on $(a,b)\cap \Bbb Q$.

Since $f_n\to f$ uniformly on  $(a,b)\cap \Bbb Q\implies |f_n(x)-f(x)|<\epsilon$ $\forall n\ge m;\forall x\in (a,b)\cap \Bbb Q$.
Also each $f_n$ is uniformly continuous.Hence $|f(x)-f(y)|\le |f(x)-f_m(x)|+|f_m(x)-f_m(y)|+|f_m(y)-f(y)|<\epsilon $ whenever $|x-y|<\delta$.
So $f$ is uniformly continuous on $(a,b)\cap \Bbb Q$ and hence can be continuously extended to $(a,b)$ and hence to $[a,b]$.

Now $\sup _{x\in (a,b)\cap \Bbb Q} |f_n-f|<\epsilon$ forall $\epsilon>0$.
Hence $|f_n(r)-f(r)|=|f_n(\lim c_n)-\lim f(c_n)|=\lim |(f_n-f)(c_n)|<\epsilon$
Hence $\sup _{x\in (a,b)\cap \Bbb Q^c} |f_n-f|<\epsilon$ forall $\epsilon>0$.
So $f_n\to f$ uniformly on $[a,b]$
