Irrationals Dense in Reals
Let $\R\setminus\Q$ be the set of irrational numbers.
Then $\R\setminus\Q$ is everywhere dense in $T$.
Let $x \in \R$.
Let $U \subseteq \R$ be an open set of $T$ such that $x \in U$.
From Between Every Two Reals Exists a Rational, $\exists p\in\Q:p\in (x . . x+\epsilon)$
Thus, we can define the open interval $V_1=(x . . p)\subseteq V_0$.
Similarly, we see $\exists q\in\Q:q\in (x . . p)$.
We can then define an open interval $V_2=(q. . p)\subseteq V_1$.
We have $V_2\subseteq V_1$, $V_1\subseteq V_0$ and $V_0\subseteq U$.
By successively applying Subsets Transitive, it follows that $V_2 \subseteq U$.
Note that $x \notin V_2$, since $x<q<p<x+\epsilon$.
From Between Every Two Rationals Exists an Irrational, there exists $y \in \R\setminus\Q: y \in (p. . q)=V_2$.
As $x \notin V_2$, it must be the case that $x \ne y$.
Since $V_2\subseteq U$, $U$ is an open set of $T$ containing $x$ which also contains an element of $\R\setminus\Q$ other than $x$.
As $U$ is arbitrary, it follows that every open set of $T$ containing $x$ also contains an element of $\R\setminus\Q$ other than $x$.
That is, $x$ is by definition a limit point of $\R\setminus\Q$.
As $x$ is arbitrary, it follows that all elements of $\R$ are limit points of $\R\setminus\Q$.
The result follows from the definition of everywhere dense.