A fellow member of the community asked: "there isn't a sequence of continuous function on $[0,1]$ that converges pointwise to the function $f$ on $[0,1]$ defined by $f(x)=0$ if $x$ is rational and $f(x)=1$ if $x$ is irrational." This is the reverse of $\chi_{\mathbb{Q}}$, but the proof is obviously the same.
I'm wondering about a solution that only uses the definition of continuity and point-wise convergence. I answered the OP with the following attempt:
Attempt 1: " Suppose that $\{f_{n}\}_{n=1}^{\infty}$ is a point-wise convergent sequence that converges to $f$. Let $x \in [0,1]$ and WLOG, suppose that $x$ is irrational.
Let $\epsilon>0$. WLOG, suppose that $\epsilon<1$.
1) Then there exists $N_{1} \in \mathbb{N}$ such that $b \in \mathbb{N}$ and $b>N_{1}$ imply that $|f_{b}(x)-f(x)|=|f_{b}(x)-1|<\frac{\epsilon}{2}$.
2) Since $\{f_{n}\}_{n=1}^{\infty}$ is a sequence of continuous functions, there exists some $\delta>0$ such that $|x-y|<\delta$ implies that $|f_{b}(y)-f_{b}(x)|<\epsilon$ for each $y \in [0,1]$. Let $y \in [0,1]$ such that $|x-y|<\delta$. Then choose $p$ such that $p \in \mathbb{Q}$, where $x<p<y$.
4) Since $\{f_{n}\}_{n=1}^{\infty}$ is pointwise convergent, there exists $N_{2} \in \mathbb{N}$ such that $a \in \mathbb{N}$ and $a>N_{2}$ imply that $|f_{a}(p)-f(p)|=|f_{a}(p)|<\frac{\epsilon}{2}$.
5) Let $N=\max\{N_{1},N_{2}\}$. Suppose that $n>N$. By hypothesis, $|f_{n}(p)|<\frac{\epsilon}{2}$ and $|f_{n}(x)-1|<\frac{\epsilon}{2}$. However, by (3), we know that $|f_{n}(x)-f_{n}(p)|<\epsilon$. But this is clearly a contradiction."
However, this doesn't work, since it assumes that the same $\delta$ will characterize continuity for any $f_{n}$ where $n>N$.
**as a side-remark, is there any literature on the idea of a single $\delta$ working to describe continuity in a sequence of point-wise convergent functions $f_{n}$ for sufficiently large $n$?
Attempt 3: sigh, here is another proof that ultimately rests on baire category for contradiction:
Let $\{f_n(x)\}$ be a sequence of continuous functions that converge pointwise.
We show that $A=\{x \in \mathbb{R} \mid f_n(x) \textrm{converges}\}$ is a Borel set [and more specifically, $G_\delta$.
If $x \in A$, then there exists $N \in \mathbb{N}$ so that $n,m \geq N$ implies that $V_{x,n,j}=|f_n(x)-f_m(x)|<\frac{1}{j}$ for each $j \in \mathbb{N}$. Then Let $V_{n,j}=\bigcup V_{x,n,j}$. Since $V_{n,j}$ is open, let the open set $U_{n,j}$ be defined by $U_{N,J}=F^{-1}(V_{n,j})$. But then $$\bigcap_{j \in \mathbb{N}} \bigcup_{N \in \mathbb{N}} \bigcap_{n,m \geq N} U_{n,j}=A$$ is $G_\delta$
Then use Baire category to show that $\mathbb{Q}$ is not $G_\delta$, and so the result follows.
I'm still interested in a proof that does not use this, and also in a possible proof verification for my proof given in the answers section.