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!
 A: Here is yet another proof:
Let $(X,d)$ be a metric space and let $F,G\subseteq_{\text{cl.}}X$ with $F\cap G=\emptyset$. Then $F\subseteq(X\setminus G)\subseteq_{\text{op.}}X$ and $G\subseteq(X\setminus F)\subseteq_{\text{op.}}X$. Thus, for each $x\in F$ and $z\in G$, $\exists\epsilon_x,\epsilon_z>0$ such that $B_{\epsilon_x}(x)\subseteq(X\setminus G)$ and $B_{\epsilon_z}(z)\subseteq(X\setminus F)$.
Next, let $$U=\bigcup_{x\in F}B_{\frac{\epsilon_x}2}(x)\subseteq_{\text{op.}}X\quad\text{and}\quad V=\bigcup_{z\in G}B_{\frac{\epsilon_z}2}(z)\subseteq_{\text{op.}}X$$ and assume to the contrary that $\exists y\in U\cap V$. Then $\exists x\in F$ and $\exists z\in G$ such that $y\in B_{\frac{\epsilon_x}2}(x)$ and $y\in B_{\frac{\epsilon_z}2}(z)$. Assume without loss of generality that $\epsilon_x\geq\epsilon_z$. Then, by the triangle inequality (T.I.),$$d(x,z)\overset{\text{T.I.}}{\leq}d(x,y)+d(y,z)<\frac{\epsilon_x}2+\frac{\epsilon_z}2\leq\frac{\epsilon_x}2+\frac{\epsilon_x}2=\epsilon_x$$ and thus $z\in B_{\epsilon_x}(x)\subseteq(X\setminus G)$, contrary to $z\in G$ (if $\epsilon_x\leq\epsilon_z$, a similar argument shows that $x\in B_{\epsilon_z}(z)\subseteq(X\setminus F)$, contrary to $x\in F$). Therefore, $U\cap V=\emptyset$ and so, since $F,G\subseteq_{\text{cl.}}X$ with $F\cap G=\emptyset$ were arbitrary, $(X,d)$ is normal.
A: 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.
