A $T_1$ space with a compatible normal sequence of open covers is metrizable

Put $V_n=\{(x,y):y\in\operatorname{St}(x,\mathcal U_n)\}$. Each $V_n$ is symmetric, contains the diagonal, and normality gives $V_{n+1}\circ V_{n+1}\subseteq V_n$: two successive $\mathcal U_{n+1}$-links lie in the star of one member, which is contained in a member of $\mathcal U_n$.

Define $d(x,y)$ as the smaller of $1$ and the infimum of $\sum_{r=1}^k2^{-n_r}$ over finite chains $x=x_0,\ldots,x_k=y$ with $(x_{r-1},x_r)\in V_{n_r}$, taking the infimum of an empty collection to be $+\infty$. Reversing a chain gives symmetry. Concatenation gives the triangle inequality within a chain-connected component, while points in different components have distance $1$; truncation at $1$ preserves the triangle inequality. The diagonal chains give $d(x,x)=0$.

The containment in step 1.1 lets every chain of total weight below $2^{-n-1}$ be compressed, from its finest links upward, to a $V_n$-link. Hence $d(x,y)<2^{-n-1}$ implies $(x,y)\in V_n$; conversely $(x,y)\in V_n$ gives $d(x,y)\le2^{-n}$.

Compatibility (i) and step 3.1 show $d(x,y)>0$ when $x\ne y$. Compatibility (ii), the two bounds in step 3.1, and [L1] show that the $d$-balls and the original neighbourhoods contain one another at every point. Thus $d$ is a metric inducing the given topology.