Alternative proof: show that any metrizable space $X$ is normal - Part 2 This is a follow up to one of my earlier questions
I am reading some stuff online and saw a proof as follows



Per a comment in Part 1 in linked, We know that $d(C_1,C_2)$ could easily be zero for disjoint closed sets (Intuition: Think Gabriel's Horn). Does the union of all the $B_{\frac{d}{3}}(x)$ balls in the proof make sense?
 A: No, it does not work.  However, it can be made to work with the following adjustments.  Put
\begin{align*}
U_1&=\bigcup\limits_{x\in C_1}B_{\frac{d(x,C_2)}{3}}(x), \\
U_2&=\bigcup\limits_{x\in C_2}B_{\frac{d(x,C_1)}{3}}(x).
\end{align*}
Now if $y\in U_1\cap U_2$, there exists $x_1\in C_1, x_2\in C_2$ such that $y\in B_{\frac{d(x_1,C_2)}{3}}(x_1)\cap B_{\frac{d(x_2,C_1)}{3}}(x_2). $  Then
\begin{align*}
d(x_1,y)&<\frac{d(x_1,C_2)}{3}, \\
d(x_2,y)&<\frac{d(x_2,C_1)}{3},
\end{align*}
so that 
$$ d(x_1,x_2)<\frac{d(x_1,C_2)}{3}+\frac{d(x_2,C_1)}{3}. $$
But since $d(x_1,C_2)\leq d(x_1,x_2)$ and $d(x_1,C_2)\leq d(x_1,x_2)$, this implies that
\begin{align*}
2d(x_1,C_2)&<d(x_2,C_1), \\
2d(x_2,C_1)&<d(x_1,C_2),
\end{align*}
an obvious contradiction.  Thus $U_1\cap U_2=\varnothing$.
A: As you suspected, the possibility of $d(C_1,C_2)=0$ messes up this proof. For a correct proof, define $U_1=\{x:d(x,C_1)<d(x,C_2)\}$ and $U_2=\{x:d(x,C_2)<d(x,C_1)\}$.  You'll need that $d(x,C_i)=0$ if and only if $x\in C_i$ (because $C_i$ is closed) and that $d(x,C_i)$ is a continuous function of $x$ (because of the triangle inequality).
