Setep 1: $[\Bbb{Q}$ is dense in $\Bbb{R}]$
Every non empty open interval interval on $\mathbb{ R }$contains at least one $r\in\mathbb{Q }$.
Limit points of $\mathbb{ Q }$is the whole of $\mathbb{R }$.
For any $x\in \mathbb{R } $ there exists a sequence $(r_n) _{n\in\mathbb{N}} \subset \mathbb{Q}$ such that $(r_n) \to \mathbb{R}$.
Give any $x\in \mathbb{R }$ there exists a rational number $r$ such that $d(x, r)<\epsilon $ for all $\epsilon >0$
$\mathbb{Q }$ is dense in $\mathbb{R }$
All the above statements are equivalent.
It is not difficult to prove $Q$ is dense in $\mathbb{R}$.
I am going to prove using property $1)$ i.e any non-empty open interval contains at least one rational number. (In fact it contains infinitely many of them).
Choose an open interval $(a, b) \subset\mathbb{R } ,a<b $
Then, $\ell(a, b) =b-a>0$
Then by Archimedean Property, $\exists n\in {\mathbb{N}}$ such that $$\frac{1}{n}<(b-a) $$
$$\implies n(b-a) >1$$
Consider the open interval $(na, nb) $
And $$\ell(na, nb) >1$$
$\implies (na, nb) $ contains an integer. Because,
Consider, $$k=[na]=floor(na) $$
Then, $k\le na <k+1$ , $k\in \mathbb{Z}$
Claim: $k+1<nb$
Suppose, $k+1\ge nb$
Then, $[na, nb]\subseteq [k, k+1]$
$\implies \ell[na, nb]\le \ell[k, k+1]$
$\implies n(b-a) \le 1$
But this contradict that $n(b-a) >1$.
Hence, $k+1<nb$
and set, $k+1=m$
And that implies $na<m<nb$
$\implies a< \frac{m}{n}<b$
Hence, $\frac{m}{n} \in (a, b) $ and $\frac{m}{n} \in \mathbb{Q}$
So we proved any non empty open interval contains a rational number.
Hence, $\mathbb{Q}$ is dense in $\mathbb{R}$.
One can start from Step 2
Step 2:$[\Bbb{R\setminus Q}$ is dense in $\mathbb{R}]$
Let, $a, b\in \Bbb{R}$ with $a<b$
Then, the open interval$$(a-\sqrt2 , b-\sqrt2) $$ contains at least one rational number $(\text{ by density of } \Bbb{Q})$, say $r$.
So, \begin{align}& r\in (a-\sqrt2 , b-\sqrt2) \\ &\implies r+\sqrt2\in(a, b) \end{align}
And, $r+\sqrt2 \in {\Bbb{R\setminus Q}}$
Since, $(a, b) $ is any arbitrary open interval, hence every open intervals contains at least one irrational, implies irrationals are dense in Real.
This is a standard textbook proof. You can find it any Real analysis book.