TITLE: Proof that every metric space is normal
QUESTION [7 upvotes]: 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!

REPLY [3 votes]: 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.