Show $f$ is discontinuous at every $x_{0}$ in $\mathbb{R}$. Let $f(x)=1$ for rational numbers x, and $f(x)=0$ for irrational numbers. Show $f$ is discontinuous at every $x_{0}$ in $\mathbb{R}$.
I started with the negation of $f$'s discontinuity first.
Proof: Suppose $f$ is continuous at some $x_{0}$ in $\mathbb{R}$.
Then, for all $\varepsilon > 0 $, there exists $\delta > 0$, s.t for all $x$ in $\mathbb{R}$ that satisfies $|x-x_{0}| < \delta$, we have $|f(x)-f(x_{0})| < \varepsilon$.
Let $\varepsilon=1/2$.
We know that for any $B_{\delta}(x_{0})$ for all $x_{0}$ in $\mathbb{R}$, we always can achieve some rational number and irrational number.
Then, for all $x_{0}$ we can always get, $|f(x)-f(x_{0})|=|1-0|=1$.
However, $1>1/2$ is a contradiction.
Hence,   $f$ is discontinuous at every $x_{0}$ in $\mathbb{R}$.
$$\blacksquare$$
Well, it's a very well-known problem, but I did not want to use sequence $(x_{n})$ that converges to $x_{0}$ because not using it seems a little bit easier for me to understand.
Does my proof make sense? Is there a simpler, or shorter approach for this kind of problem?

Here it is another way to approach it for the sake of curiosity.
Suppose that $f$ is continuous at $q\in\mathbb{Q}$.
Then for every sequence $q_{n}$ which converges to $q$, the sequence $f(q_{n})$ must converge to $f(q) = 1$.
Based on such assumption, we shall consider the following sequences:
\begin{align*}
\begin{cases}
x_{n} = q + \dfrac{1}{n}\\\\
y_{n} = q + \dfrac{\sqrt{2}}{n}
\end{cases}
\end{align*}
Notice that both $x_{n}$ and $y_{n}$ converges to $q$, but $x_{n}$ is rational and $y_{n}$ is irrational.
Consequently, we have that $f(x_{n}) = 1$ and $f(y_{n}) = 0$, which contradicts our primary assumption.
Similar reasoning applies to the irrational values. Thus $f$ is everywhere discontinuous.
