Equivalence of Forms of Baire Category Theorem

Question

I am trying to show the equivalence of two forms of the Baire Category Theorem. These are the two statements:

1. Let $(X,d)$ be a complete metric space. Let $U_n$ be a dense, open set for each $n \in \Bbb N$. Then $\bigcap_{n=1}^\infty U_n$ is dense (in $X$).
2. Let $(X,d)$ be a complete, non-empty metric space. If $X = \bigcup_{n=1}^\infty A_n$, then there exists $N \in \Bbb N$ such that $\text{Int} \bar{A_N} \neq \emptyset$.

I can show that $1.$ implies $2.$, and for the case that $X = \emptyset$ that $2.$ implies $1.$. My issue is showing that, for non-empt $X$, $2.$ implies $1.$. I've been trying various different manipulations of the closure of $\bigcap_{n=1}^\infty U_n$, eg using de Morgan's law, but I can't get anywhere.

Any advice would be most gratefully received! Thanks! :)

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UPDATE: I now have a solution - see answer, if you are interested.

Answer 1

I now have a solution to this - credit to my lecturer, the excellent Dr Zsak.

Observe that $\bigcap_{n=1}^\infty U_n \neq \emptyset$. Indeed, if $\bigcap_{n=1}^\infty U_n = \emptyset$, then we have $$X = \emptyset^c = \left( \bigcap_{n=1}^\infty U_n \right)^c = \bigcup_{n=1}^\infty U_n^c,$$ but $\text{Int}(U_n^c) = \emptyset$ for all $n \in \Bbb N$, contradicting our assumption.

Now, let $Y = B(x,r)$ for some $x \in X$ and $r > 0$. Observe that $Y \cap U_n$ is open and dense in $Y$ (not $X$). By above, $$Y \cap \bigcap_{n=1}^\infty U_n = \bigcap_{n=1}^\infty (Y \cap U_n) \neq \emptyset.$$ Thus, $\bigcap_{n=1}^\infty U_n$ meets every open ball, and hence is dense.