If A is dense in (X,T) and A = the countable intersection of open sets in X, then each open set is dense

If $A$ is dense in $(X,T)$ and $A$ is an $G_{\delta}$ set $A$ $= \bigcap _{n=1}^{\infty} B_n$ such that $B_n \in T$ $\Rightarrow B_n$ for each n is dense in $(X,T)$.

I have: $A$ $= \bigcap _{n=1}^{\infty} B_n$ $\Rightarrow$ $A$ $\subseteq \bigcap _{n=1}^{\infty} B_n$ $\Rightarrow \forall n \in \Bbb N, A \subseteq B_n$.

And any subset that contains a dense subset in X is dense in X.

I feel like I'm skipping something between the subset and the for all statements. Does this proof make sense or should I have more between the two statements?

• It really is that easy. To say it a little more concisely, you need only observe that if $U$ is a non-empty open set, then for each $n\in\Bbb N$ we have $U\cap B_n\supseteq U\cap A\ne\varnothing$, so $B_n$ is dense in $X$. – Brian M. Scott May 29 '16 at 17:15

You can conclude directly from $A = \bigcap_{n=1}^\infty B_n$ that $A \subseteq B_n$ for all $n$.
And so $X = \overline{A} \subseteq \overline{B_n} \subseteq X$ so $B_n$ is dense (expanding the comment about "any subset that contains a dense subset of $X$ is dense in $X$").