Urysohn Metrization Theorem
Solution 1:
I think I got it.
Given $x\in U_k$ for some $U_k$ in the base, since $X$ is normal, $\{x\}$ is closed and there exists $$\{x\}\subset U_l \subset \overline {U_l} \subset U_k,$$ and we have $(l,k)\in A$.
Now let $\epsilon = \frac{1}{2^{l+k}}$, if $\rho(x,y) <\epsilon$, then we have $$\epsilon > \rho(x,y) = \sum_{(n,m)\in A} \frac{1}{2^{n+m}} |f_{n,m}(x) - f_{n,m}(y)| \geq \frac{1}{2^{l+k}} |f_{l,k}(x) - f_{l,k}(y)|$$ this implies $$|f_{l,k}(x) - f_{l,k}(y)| = |0-f_{l,k}(y)| = f_{l,k}(y) <1$$ from construction, we must have $y\in U_k$.