Let $f_n \in C([0,1])$ be a sequence of functions with $f_n(x) \to f(x)$ for all $x$. Consider the set of discontiuities of $f$, which I denote by $\mathcal{D}$. Then, we can define:
$$\mathcal{D} = \bigcup_N C_N, \text{ where } C_N =\{x : \exists x_k \to x \; \text{s.t. for all } k, \; |f(x_k) - f(x)| \ge \frac{1}{N} \}$$
I attempted to prove that $C_N$ is closed as follows. Suppose that $y_m \in C_N$ with $y_m \to y$ in $[0,1]$. Then, by definition, there exists $z_{km} \to y_m$ with $|f(z_{km}) - f(y_m)| \ge \frac{1}{N} \; \forall k$. We observe that $z_{kk} \to y$, so we consider:
$$|f(y) - f(z_{kk})|$$
$$\ge |f(y_k) - f(z_{kk})| - |f(y) - f(y_k)|$$$$\ge \frac{1}{N} - |f(y) - f_n(y)| - |f_n(y) - f_n(y_k)| - |f_n(y_k) - f(y_k)|$$
Now, naturally, it makes sense to let $n \to \infty$ so that $|f(y) - f_n(y)|, |f(y_k) - f_n(y_k)| \to 0$. However, it seems to me that we do not necessarily have that $|f_n(y) - f_n(y_k)| \to 0$, and indeed I found counterexamples for this. Therefore, instead I considered that:
$$\mathcal{D} = \bigcup_N \hat{C}_N, \text{ where } \hat{C}_N =\{x : \exists x_k \to x \text{ s.t. } \limsup_k |f(x_k) - f(x)| \ge \frac{1}{N} \}$$
Now we can continue the proof with:
$$ \limsup_k |f(y) - f(z_{kk})| \ge \limsup_k \bigg(\frac{1}{N} - |f(y) - f_n(y)| - |f_n(y) - f_n(y_k)| - |f_n(y_k) - f(y_k)| \bigg)$$
$$\ge \frac{1}{N} - |f(y) - f_n(y)| - \liminf_k |f_n(y) - f_n(y_k)| - \liminf_k |f_n(y_k) - f(y_k)|$$
$$\ge \frac{1}{N} - |f(y) - f_n(y)| - \liminf_k |f_n(y_k) - f(y_k)| $$
(since each $f_n$ is continuous.)
$$\ge \frac{1}{N} - |f(y) - f_n(y)| - |f_n(y_1) - f(y_1)| $$
$$\ge \frac{1}{N}$$
(since $f_n(x) \to f(x)$ for all $x$) as desired. Therefore, $\hat{C}_N$ is closed for each $N$.
I am satisfied with this proof, but I would find it quite unintuitive if $\hat{C}_N$ is always closed, as proven, but $C_N$ is not in general. Is there a way to extend this proof, or use another method, to deduce that $C_N$ is closed, or does there exist a counterexample which proves it is not true in general?
(This is part of an attempt to prove a result using the Baire category theorem; please do not provide hints on the broader result, which I am still attempting.)