Alternative proof: show that any metrizable space $X$ is normal - Part 1

Question

There is a proof online that shows that all metric spaces are normal. The proof is as follows

However, it has the additional baggage of needing to show that $d(x,A)$ is continuous and $U,V$ are open.

Can we do make it even simpler using the following:

• Let $C, D$ be disjoint closed sets in $X$, $d$ be the metric that generates topology on $X$.

• Define the set to set distance $d(C,D) = \inf\{d(c,d)| c \in C, d \in D\}$.

• Then take $r = d(C,D). r > 0$ since $C,D$ are disjoint.

• Let $U = \bigcup_{x \in C} B_{\frac{r}{2}}(x), V = \bigcup_{y \in D} B_{\frac{r}{2}}(y)$ are disjoint open sets containing $C, D$
  respectively.

Is this good enough as an alternative proof?

Answer 1

It’s not necessarily true that $d(C,D)>0$ for disjoint closed sets $C$ and $D$. As a counterexample, take $C$ to be the $x$-axis in $\Bbb R^2$ and $D$ to be the graph of the hyperbola $xy=1$; then $C$ and $D$ are closed, $C\cap D=\varnothing$, and $d(C,D)=0$.