real analysis - Is this proof ok?

My friend and I were trying to prove that a function can't be discontinuous on the irrationals. His proof was this:

Let $f:\mathbb{R}\to\mathbb{R}$ be continuous on all of $\mathbb{Q}$. Then for any $\{x_n\}\subset\mathbb{Q}$ with $x_n\to r\in\mathbb{Q}$, we have $\lim_{n\to\infty}f(x_n)=f(r)$. Now choose $\{x_n\}$ such that $\{x_n\}$ is Cauchy. Then for any $\epsilon>0$ there exists an $N_0\in\mathbb{N}$ such that for all $n,\ m\ge N_0$, $|x_m-x_n|<\epsilon$.

Now in the interval $(x_n,\ x_m)$ there's an irrational $\gamma$ between them. Therefore, $\gamma$ also satisfies the definition of continuity, and so $f(x)$ is continuous for at least one $\gamma\in\mathbb{Q}^c$ and so it can't be continuous on only the rationals.

Something about this doesn't sit well with me, it feels so off, but I can't tell why. If it were this simple all along, what's with all the $F_\sigma$ and $G_\delta$ sets and Baire Spaces and Theorems and all this other complicated stuff?