 Solution

The complete metric space $\mathbb{R}$ supplies a counterexample: $A_n=[n,\infty )$ are descending nonempty closed sets, but no real number belongs to them all.

Now assume shrinking diameters. Choose $x_n\in A_n$. Given $\epsilon >0$, choose $N$ so that $\mathrm{diam}(A_N)<\epsilon$. Whenever $m,n\ge N$, both chosen points lie in $A_N$, so $d(x_m,x_n)<\epsilon$. Thus $(x_n)$ is a Cauchy sequence. By completeness it converges to some $x\in X$. For each fixed $N$, the tail lies in the closed set $A_N$, so its limit lies there too. Hence $x\in {\bigcap }_n{A}_n$. Any two points of this intersection have distance at most every $\mathrm{diam}(A_n)$, so their distance is zero. The shrinking intersection contains exactly one point. This is the Cantor intersection theorem.

For the dense-open conclusion, begin inside an arbitrary nonempty open set $O\subseteq X$; taking $O=X$ gives the requested existence. Since $U_1$ is a dense subset, choose $x_1\in O\cap U_1$ and $0<r_1\le 1/2$ small enough that the closed ball $A_1=\overline{B}(x_1,r_1)$ is contained in $O\cap U_1$. Here $\overline{B}$ denotes the closed ball, not an assumption about the closure of the open ball.

Inductively, the open ball $B(x_n,r_n)$ meets the dense open set $U_{n+1}$. Choose $x_{n+1}$ in that intersection and a positive radius $r_{n+1}\le 2^{-(n+1)}$ so small thatSuch a radius exists by openness and the triangle inequality. These nonempty closed sets descend and have diameters at most $2r_n\to 0$. The preceding result gives a point in every $A_n$, hence in $O$ and every $U_n$. Therefore ${\bigcap }_n{U}_n$ is nonempty; in fact it is dense in $X$. This proves the relevant Baire category theorem directly, without assuming closed balls are compact.