Limit points of rationals in $R$ Some background:
I was going through the lecture 9 of real analysis series of Harvey Mudd college. The lecture discusses open balls, neighbourhoods,limit points, isolated points and interior points. Till this lecture the series has covered construction of rationals, field axioms, dedekind cut, lub, glb, completeness,
principle of induction, well ordering principle, countable and uncountable sets.
In this lecture prof Francis Su goes over an example that relates to the limit points of $Q$ in $R$. He goes on to say that any interval on real line contains rationals.
Question:
1.) Given the background is there a way to prove that every interval on $R$ contains $Q$ without using the density property of rational nos.
2.) How can we prove that the limit points of $Q$ is the whole of $R$ without using the density property
 A: *

*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.

Question: 1) Given the background is there a way to prove that every
interval on R contains Q without using the density property of
rational nos.


Question : 2) How can we prove that the limit points of Q is the whole of R
without using the density property.

Your questions are equivalent to prove the density property of $\mathbb{Q }$ in $\mathbb{R}$.So "without density property " - this doesn't make sense. All are equivalent to density property.
But 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 $\mathbb{R}$
And all equivalent conditions are satisfied.
Correct me if there is any mistake.
Hope it helps. Thanks.
{Edit: I want to share how to think about the proof .
Step 1 : Choose an open interval $(a, b) $.
And assume that it contains a rational number say $r$.
Then, $a<r<b$ and $r=\frac{m}{n} $ where$ m, n\in \mathbb{Z}$ and $n>0$.
Then, $a<\frac{m}{n}<b$
$\implies na<m<b$
Step 2: Now the the proof of existence of a rational number reduces to the proof of the existence of an integer.
Prove the result that any interval of length strictly larger than  $1 $ contains an integer. }
