Proof that every metric space is normal

I've seen this proof of the fact that a metric space is normal multiple times but I can't understand how it's valid.

Notation used: For $x \in X$, and $Y$ a subset of $X$, define $D(x,Y)=\inf \{d(x, y): y \in Y\}$.

The above proof uses that if $Y$ is a closed set, then $D(x, Y)>0$, $\forall x \in X \setminus Y $ and this in turn implies $\space D(x, Y)> \epsilon$, $\forall x \in X \setminus Y $ for some $\epsilon > 0$.

If the above proof was true, then we could also argue that for any $y \in Y$ we have $B(y, \frac{\epsilon}{3}) \subseteq Y$ (since $d(y, x)> \epsilon, \space \forall x \in X \setminus Y \Rightarrow B(y, \frac{\epsilon}{3}) \cap \{X \setminus Y\} = \emptyset) $ so $Y$ is open which is obviously false.

I don't know what I'm missing becuase this kind of proof seems to appear everywhere.

Thank you!

The proof in the image you linked to is not a valid proof.

It's not necessarily true that for all pairs $C_1,C_2$ of nonempty disjoint closed subsets of $X$, we have $d(C_1,C_2) > 0$.

For example, if $X=\mathbb{R}^2$, and \begin{align*} C_1&= \{(a,0)\mid a\in\mathbb{R}\}\\[4pt] C_2&=\{\bigl(b,{\small{\frac{1}{b}}}\bigr)\mid b > 0\}\\[4pt] \end{align*} then $C_1,C_2$ are nonempty disjoint closed subsets of $X$, but $d(C_1,C_2)=0$.

What is true is that if $C$ is a nonempty closed subset of $X$, and $x\in X$, then $d(x,C)=0\;$if and only if $x\in C$.

Proof:$\;$If $x\in C$, then of course, $d(x,C)=0.\;$Conversely, suppose $C$ is a nonempty closed subset of $X$, and $x\in X$ is such that $d(x,C)=0.\;$Then since $d(x,C)=0$, it follows that $B(x,r)\cap C$ is nonempty, for all $r > 0,\;$hence $x$ is in the closure of $C$, which is $C$.

Hence, if $C$ is a nonempty closed subset of $X$, then for all $x\in X{\setminus}C$, we have $d(x,C) > 0$.

The proof can then be continued as follows . . .

• For each $x\in C_1$, let $r={\large{\frac{d(x,C_2)}{3}}}$, and let $U_x=B(x,r)$.$\\[4pt]$
• For each $y\in C_2$, let $s={\large{\frac{d(y,C_1)}{3}}}$, and let $V_y=B(y,s)$.

Now let \begin{align*} U&=\bigcup_{x\in C_1} U_x\\[4pt] V&=\bigcup_{y\in C_2} V_y\\[4pt] \end{align*} It's clear that $U,V$ are open subsets of $X$, with $C_1\subseteq U$, and $C_2\subseteq V$.

Suppose $U\cap V\ne{\large{\varnothing}}$.

Let $z\in U\cap V$.

Since $z\in U$, we must have $z\in U_x$, for some $x\in C_1$, hence $d(x,z) < r$, where $r={\large{\frac{d(x,C_2)}{3}}}$.

Since $z\in V$, we must have $z\in V_y$, for some $y\in C_2$, hence $d(y,z) < s$, where $s={\large{\frac{d(y,C_1)}{3}}}$.

Without loss of generality, assume $r\ge s$.$\;$Then

$$3r=d(x,C_2)\le d(x,y)\le d(x,z)+d(y,z)< r+s\le 2r$$ contradiction.

Therefore $U\cap V={\large{\varnothing}}$.

It follows that $X$ is normal.