Is this proof that $U_n$ is dense correct?

Question

{Showing that Show $U_n$ is also dense.}

Since by {Second.} with $X = \mathbb{R}$, we know that upon fixing $n,$ that every point of continuity of $f$ belongs to $U_n$ and since $f$ is assumed to be continuous at all rational points, then $ U_n$ contains $\mathbb Q$. And since $\mathbb{Q}$ is dense in $\mathbb{R}$(by Advanced Calculus), hence $U_n$ is dense in $\mathbb{R}.$ \

Where Second. is "Showing that $\bigcap_n U_n$ is precisely the set of points at which $f$ is continuous."

Is this proof correct? I am confused about directly saying that $U_{n}$ is dense in $\mathbb{R}$ because $\mathbb{Q}$ is dense in $\mathbb{R}$. Is there details that should be added?

Answer 1

You could put some more details into it, for example this way:

If $A\subseteq B$, then $\bar{A}\subseteq\bar{B}$. Now, choose $A=\mathbb{Q}$ and $B=U_n$. Since $\bar{\mathbb{Q}}=\mathbb{R}$, it follows that $\mathbb{R}\subseteq \bar{U_n}\subseteq\mathbb{R}$.

The idea is correct, though. If a set contains a dense set, then it is dense itself, just as you claimed.