Limit of a pointwise sequence of continuous functions Let $(f_n)$ be a sequence of continuous functions on $[0,1]$ converges pointwise everywhere. Let $f$ be the limit function on $[0,1]$. Show that $\{t\in [0,1]:f\mbox{ is continuous at }t\}\neq\emptyset$.
I tried to solve by the following method.
Let $F_n=\{x\in [0,1]:O(f,x)\geq 1/n\}$. It is clear that $F_n$ is closed. If we can show that it has non empty interior then by Baire Category Theorem the desired set is dense in $[0,1]$. To prove that $F_n$ has empty interior I am facing problem.
 A: Let $A(\epsilon, N) = \{x \in [0,1]:\; \forall n,m\ge N, \; |f_n(x) - f_m(x)| \le \epsilon \}$.  This is closed, and $\bigcup_N A(\epsilon, N) = [0,1]$.  By the Baire Category Theorem, some $A(\epsilon, N)$ has nonempty interior.  Moreover, by 
repeating this argument with $[0,1]$ replaced by a closed interval in the interior of $A(\epsilon, N)$ and $\epsilon$ replaced by $\epsilon/2$, some $A(\epsilon/2, N')$ has nonempty interior in $A(\epsilon, N)$.  In this way we get a nested sequence of closed intervals $[a_k, b_k]$ and such that $[a_k, b_k]$ is contained in the interior of $A(2^{-k}, N_k)$ for some $N_k$.
Take $t$ in the intersection of these intervals. I claim $f$ is continuous at $t$.
For any $\epsilon > 0$ we can take $k$ so $\epsilon/3 > 2^{-k}$, and $\delta > 0$ so $(t-\delta, t+\delta)$ is contained in $A(2^{-k}, N_k)$, and also so that $|x - t| < \delta$ implies $|f_{N_k}(x) - f_{N_k}(t)| < 2^{-k}$.  For $n \ge N_k$ and $x \in (t-\delta, t+\delta)$ and we have $|f_n(x) - f_{N_k}(x)| < 2^{-k}$, so that
$|f(x) - f_{N_k}(x)| \le 2^{-k}$.   Thus if $|x - t| < \delta$, 
$$ |f(x) - f(t)| \le |f(x) - f_{N_k}(x)| + |f_{N_k}(x) - f_{N_k}(t)| + |f_{N_k}(t) - f(t)| < 3 \cdot 2^{-k} < \epsilon$$
