Proof verification: Let $K$ be compact, then there exists disjoint open subsets $U,V \subset X$ so that $x \in U$ and $K \subset V$ 
Let $(X,d)$ be a metric space and $K \subset X$ a compact subset.
  We define $$\text{dist}_K(x) := \inf_{y \in K} d(x,y).$$
  Show that
  
  
*
  
*for every $x \in X \setminus K$ there exists disjoint open subsets $U,V \subset X$ so that $x \in U$ and $K \subset V$. Note: I know this is very similar to this question and this one but here, I have a different metric.
  
*$\text{dist}_K(x) = 0 \iff x \in K$.

My Proof


*

*Let $x \in X \setminus K$. Then we have $\text{dist}_K(x) > 0$ and therefore $U := B\left(x, \frac{1}{2}\text{dist}_K(x)\right) \subset X$ (which is the open ball centered around $x$ with radius $ \frac{1}{2}\text{dist}_K(x)$) which is an open subset containing $x$.
Now, define $V := \bigcup_{k \in K} B\left(k, \frac{1}{2}\text{dist}_K(x)\right)$ which is open in $X$ as union of sets which are open in $X$.
Now, to show $\tilde{U} := U \cap V = \emptyset$, we assume there exists an $u \in \tilde{U}$ and seek a contradiction.
By definition we have
\begin{equation}
d(u,x) < \text{dist}_K(x) \quad \forall x \in X \setminus K
\qquad \text{and} \qquad
d(u,k) < \text{dist}_K(x) \quad \forall k \in K
\end{equation}
Adding both equations, for all $x \in X \setminus K$ and all $k \in K$ we have
\begin{equation}
\inf_{y \in K} d(x,y)
> d(x,u) + d(u,k)
\overset{\triangle \neq}{\ge} d(x,k),
\end{equation}
which is a contradiction.

*"$\implies$": 
If $\text{dist}_K(x) = 0$ we have $I := \inf_{y \in K} d(x,y) = 0$
By the definition of the infimum there exists a sequence $(x_n)_{n \in \mathbb{N}} \subset K$ with $\lim_{n \to \infty} x_n = x$.
But since $K$ is compact it's also closed and therefore we know that $x \in K$.
"$\impliedby$": If $x \in K$, we have
\begin{equation}
\text{dist}_K(x)
= \inf_{y \in K} d(x,y)
= d(x,x)
= 0,
\end{equation}
because $d$ is a norm and therefore positive definite on $X$.
 A: Your proof of $1$ isn't correct because until you prove $2$, you don't know that $d(x, K) \gt 0$.
In any event, but I'd simplify and generalize the proof so that it works in any Hausdorff space.
For each $y \in K~ \exists U_y, V_y~(y \in U_y, x \in V_y, U_y \cap V_y = \emptyset)$.  The $U_y$ are an open cover of $K$ so choose finitely many of them that cover $K$, $U_{y_1}, \ldots, U_{y_n}.$  Then $V= \bigcap_{k=1}^n V_{y_k}$ is a finite intersection of open sets, so it's open, and by construction $V \cap U_{y_k} = \emptyset$ for all $k$, so let $U= \bigcup_{k=1}^n U_{y_k}.$  Then $x \in V, U \cap V = \emptyset$ and $K \subseteq U.$
Edited to correct comment on proof in original post.
A: For every $k\in K$ consider $B(k, \frac13d(k,x))$ and define $V=\bigcup_{k\in K} B(k, \frac 13 d(k,x)).$ Define $U=B(x,\frac 13 d(x,K)).$ (Note that $d(x,K)$ is shown in $2$.)
Assume $y\in U\cap V.$ So, there exists $k\in K$ such that $y\in B(k, \frac13 d(k,x)).$ But then
$$d(x,K)\le d(x,k)\le d(x,y)+d(y,k)\le \frac 13 d(x,K)+\frac 13 d(x,K)=\frac 23d(x,K)$$ which is not possible unless $d(x,K)=0.$ Since $d(x,K)>0$ we have shown that $U\cap V=\emptyset.$
In the second case, assume that $d(x,K)=0.$ So, there is a sequence $(x_n)$ of points of $K$ such that $\inf_n d(x,x_n)=0.$ Since $K$ is compact there exists a subsequence of $(x_n)$ which converges to a point $x_0\in K.$ Let's denote the subsequence by $(y_n).$ We have that
$$d(x,x_0)\le d(x,y_n)+d(y_n,x_0).$$ Since $\inf_n d(x,x_n)=0$ and $\lim_n y_n=x_0$ we get that $d(x,x_0)=0.$ That is $x\in K.$
