Proof of pointwise convergence of a sequence of continuos functions.

Question

I've been trying to prove the following statement:

There is no sequence of continous functions $f_n: [0, 1] \rightarrow \mathbb{R}$ that converges pointwise to the function $f: [0, 1] \rightarrow \mathbb{R} $ defined by $f(x)=0 $ if $x $ is rational and $ f(x)=1$ otherwise.

My idea was to assume that such a sequence exists and then prove that, if that's the case, then one of the functions $f_n$ is discontinuous at some point, contradicting our assumption that the functions $f_n$ are all continuous. Here is exactly what I did:

Let $(f_n)$ be a sequence of continuous functions $f_n: [0, 1] \rightarrow \mathbb{R}$ that converges pointwise to $f$, which is defined above. Let $\epsilon>0$ be given. We are going to prove that some $f_n$ is discontinuous at $x=1/2$. Since $f_n$ converges pointwise to $f$, there exists $n_0 \in \mathbb{N}$ such that, if $n \geq n_0$, then $|f_n(1/2)-f(1/2)| < 1/2$. Since $1/2$ is rational, $f(1/2)=0$, and we get $|f_n(1/2)| <1/2.$ Moreover, for any $\delta>0,$ there exists some $x_0 \in (1/2-\delta, 1/2+\delta)$ that is irrational. Hence, there exists $n_1 \in \mathbb{N}$ such that, if $n \geq n_1,$ then $|f_n(x_0)-f(x_0)|=|f_n(x_0)-1|<1/4 \implies 3/4<f_n(x_0)<5/4$. Hence, for $N=\max\{n_0, n_1\}$, we get

$|f_N(1/2)-f_N(x_0)| \geq |f_N(x_0)|-|f_N(1/2)|>3/4-1/2=1/4.$

My idea was to show that, for $\epsilon=1/4$ and any $\delta>0$, it's possible to find some $x_0 \in (1/2-\delta, 1/2+\delta)$ such that $|f_N(1/2)-f_N(x_0)|\geq 1/4$, for some $N \in \mathbb{N},$ which would imply $f_N$ has a discontinuity at $x=1/2$, giving the contradiction for which we were looking. My "proof", however, isn't right since the $N$ I pick is based on $n_1$, which, in turn, is generally dependant on my choice of $\delta>0.$ With that being said, can anyone help me to prove the statement above?