Proof verification - If $f$ is continuous, then $f_{n}$ is continuous in at least one $n \in \mathbb{N}$ Given $f_{n} : [a,  b] \mapsto \mathbb{R}$ a sequence of functions that converge uniformaly to $f : [a,  b] \mapsto \mathbb{R}$
Prove or disprove: If $f$ is continuous, then $f_{n}$ is continuous in at least one $n \in \mathbb{N}$
My attempt at a proof:
Let $\epsilon > 0$ and $x_{0} \in [a,b]$ arbitrary
Since $f$ is continuous on $[a,b]$ there exists a $\delta_{\epsilon} > 0$ such that for all $x \in [a,b]$
$$ |x - x_{0}| < \delta_{\epsilon} \Longrightarrow |f(x) - f(x_{0})| < \frac{\epsilon}{3}$$
Because $f_{n}$ converges uniformaly to $f$ we can find an $N_{\epsilon} \in \mathbb{N}$ such that for all $n \geq N_{\epsilon}$ and all $x \in [a,b]$
$$|f_{n}(x) - f(x)| < \frac{\epsilon}{3} $$
which means that for $n \geq N_{\epsilon}$ : if $|x - x_{0}| < \delta_{\epsilon}$ then
$$ |f_{n}(x) - f_{n}(x_{0})| \leq |f_{n}(x) - f(x)| + |f(x) - f(x_{0})| + |f(x_{0}) -f_{n}(x_{0})| < \epsilon $$
Therefore $f_{n}$ is continuous for at least one $n$
Is this proof correct? Thanks.
 A: No, it is not correct and it could not possibly be correct, since the statement is false. For each $n\in\mathbb N$, define $f_n\colon[-1,1]\longrightarrow\mathbb R$ by$$f_n(x)=\begin{cases}\frac1n&\text{ if }x>0\\0&\text{ otherwise.}\end{cases}$$Then $(f_n)_{n\in\mathbb N}$ converges uniformly to the null function, but no $f_n$ is continuous.
Concerning your proof, your goal was to prove that\begin{multline}(\exists n\in\mathbb{N})(\forall x_0\in[a,b])(\forall\varepsilon>0)(\exists\delta>0):x\in[x_0-\delta,x_0+\delta]\cap[a,b]\implies\\\implies\bigl\lvert f_n(x)-f_n(x_0\bigr\rvert<\varepsilon,\end{multline}but what you actually proved was that\begin{multline}(\forall x_0\in[a,b])(\forall\varepsilon>0)(\exists n\in\mathbb{N})(\exists\delta>0):x\in[x_0-\delta,x_0+\delta]\cap[a,b]\implies\\\implies\bigl\lvert f_n(x)-f_n(x_0\bigr\rvert<\varepsilon.\end{multline}
A: Another counterexample:
For each $n \in \mathbb N$ define $f_n: [a,b] \to \mathbb R$ by 
$$f_n(x)=\begin{cases}\frac1n&\text{ if }x \in \mathbb Q \\0&\text{ otherwise.}\end{cases}$$
Given $n \in \mathbb N$ and $x \in [a,b]$ we may show that $f_n$ does not satisfy the definition of a continuous function by selecting $ \varepsilon=\frac{1}{n+1}$. So if $\delta>0$, then there is $\hat{x}$ such that $|\hat x - x|<\delta$ but $|\,f_n(\hat x) - f_n(x)|\geq \varepsilon$.
Call the limit function $f(x)=\lim_{n \to \infty}  f_n(x)$ a degenerate case of a Dirichlet function (kind of a bad joke).
