= 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 $\varepsilon>0$, choose $N$ so that $\operatorname{diam}(A_N)<\varepsilon$. Whenever $m,n\geq N$, both chosen points lie in $A_N$, so $d(x_m,x_n)<\varepsilon$. 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 $x\in\bigcap_nA_n$. Any two points of this intersection have distance at most every $\operatorname{diam}(A_n)$, so their distance is zero. \b[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 $0<r_1\leq1/2$ 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 $r_{n+1}\leq2^{-(n+1)}$ so small that
$$
\overline B(x_{n+1},r_{n+1})\subseteq B(x_n,r_n)\cap U_{n+1}.
$$
Such a radius exists by openness and the <triangle inequality>. These nonempty <closed sets> descend and have <diameters> at most $2r_n\to0$. The preceding result gives a point in every $A_n$, hence in $O$ and every $U_n$. Therefore \b[$\bigcap_nU_n$ is nonempty; in fact it is dense in $X$]. This proves the relevant <Baire category theorem> directly, without assuming closed balls are compact.
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