We have $$f(x) = \begin{cases} 1, & x = \frac{1}{n}, n\in\Bbb Z \\ 0, & elsewhere \end{cases} $$
My proof is as follows.
For the sake of contradiction, assume $f(x)$ is continuous at $x= 0$. By the Sequential Continuity Theorem, for a sequence $x_n$, if $\lim_{n \to \infty} x_n = L$ and $f(x)$ is continuous at $x = L$, then $\lim_{n \to \infty} f(x_n) = f(L)$.
Let $\{x_n\}$ be a sequence such that $x_n = \frac{1}{n}$, $n = 1,2,3,...$. We have $\lim_{n \to \infty} x_n = \lim_{n \to \infty} \frac{1}{n} = 0$. Since we assumed $f(x)$ is continuous at $x = 0$, by SCT, we have $\lim_{n \to \infty} f(x_n) = f(0)$.
The LHS evalutes to $1$, since $f(x_n) = f(\frac{1}{n})$. The RHS evalutes to $0$, since no choice of $n$ can satisfy $\frac{1}{n} = 0$. So, $1 = 0$, a contradiction. As such, we conclude $f(x)$ cannot be continuous at $x = 0$.
Is this proof correct? Also, a second part to this question tells us that $f(x)$ cannot be made continuous by altering $f(0)$ and asks to prove this claim. I'm not sure why this is true, wouldn't alterning $f(0)$ to $1$ make it continuous?
The first proof has some error. It is that $f(1/n)=1\rightarrow 1$, so $f(0)=1$, but $f(0)=0$, then this is a contradiction.
Now turn the second part.
Assume that $f(0)$ is defined as any other number, we still have $f(1/n)=1\rightarrow 1$, so if it were continuous at $x=0$, then we must have $f(0)=1$. But now $f(q_{n})=0$ for a sequence $(q_{n})$ of irrational numbers such that $q_{n}\rightarrow 0$, then $f(0)=\lim_{n}f(q_{n})=0$, once again we get $1=0$, a contradiction.