Alternative proof: show that any metrizable space $X$ is normal - Part 2

Question

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?

Answer 1

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 $U_1\cap U_2=\varnothing$.

Answer 2

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).