Let $f_n$ sequence of continuous functions on $[a,b]$ and $\lim_{n\to \infty}f_n(x)=f(x)$ uniformly on $(a,b)$. Show that $f_n\mapsto f$ uniformly on $[a,b]$.
I would like to know how to continue my proof, please.
First of all, as $f_n$ are continuous on closed interval, this sequence is uniformly continuous on $[a,b]$. By definition we have: $\forall \epsilon>0 \ \exists \delta_1>0 \ \forall x,y \in [a,b]$: $|x-y|<\delta_1 \implies|f_n(x)-f_n(y)|<\epsilon$
As $f_n$ converges uniformly to $f$ on $(a,b)$, $f$ is uniformly continuous on $(a,b)$. Moreover, the uniform continuity of $f$ on $(a,b)$ implies that $\lim_{x\to a^+}f(x)$ and $\lim_{x\to b^-}f(x)$ exist. So by extension by continuity we have that $f$ is uniofmrly continuous on $[a,b]$: $\forall \epsilon>0 \ \exists \delta_2>0 \ \forall x,y \in [a,b]$: $|x-y|<\delta_2 \implies |f(x)-f(y)|<\epsilon$
We would like to show the following: $\forall \epsilon>0 \ \exists N \ \forall n\ge N \ \forall x \in [a,b]$: $|f_n(x)-f(x)|<\epsilon$.
How can i conclude with all the hypothesises that I've done until now?
Edit: Probably I could have tried to prove by absurd and work with sequence $x_n$ which is bounded so by Bolzano-Weierstrass there is a converging sub-sequence(So cauchy) then use the uniform continuity to get a contradiction
By the uniform Cauchy criterion, there exists $N(\epsilon)$ such that for all $m > n \geqslant N(\epsilon)$ and all $x \in (a,b)$, we have $|f_m(x) - f_n(x) | < \epsilon$.
By continuity, this implies that $|f_m(a) - f_n(a)| = \lim_{x \to a+}|f_m(x) - f_n(x)| \leqslant \epsilon$, and similarly as $x \to b-$.