Proof verification for "Every infinite set $E$ in a compact set $K$ contains a limit point." Let $K$ be a compact set and $E\subset K$ be an infinite subset of $K$. Then it is to be proven that $E$ has a limit point in $K$. Let $d:K\times K\to \mathbb R^+$ be the metric in $K$.
I have tried to prove this as follows: 
I'll prove: If $E$ does not have a limit point in $K$, then $K$ is not compact. Then its contrapositive will prove the statement that I am trying to prove. 
If $E$ does not have a limit point in $K$ then for every $k\in K$, there exists a $\delta_k\gt 0$ such that the set $V_k=\{x\in K:x\ne k\text{  and } d(x,k)\lt \delta_k \}$ is such that $V_k\cap E=\emptyset$.
It follows that $E\subset V_k^c$. Therefore, $\color {red}{E\subset \cap_{k\in K}\left( V_k^c  \right)}$,  which is non-empty*. (It is to be noted here that $V_k$ is open $\implies V_k^c$ is closed and any closed subset of compact set $K$ is compact $\implies V_k^c$ is compact.) 
Also noting that $K\subset \cup_{k\in K}(V_k)$, it follows that $\cup_{k\in K}(V_k)$ can't have a finite subcover. (because if it did then there would exist finitely many indices $k_1,k_2,\cdots, k_n$ such that $K\subset \cup_{i}^n(V_{k_i})\implies E\subset \cup_{i}^n(V_{k_i})$ then for any $k\in E$, there exists some $i$ such that $k\ne k_i$ and $d(k,k_i)\lt \delta_{k_i}$, which violates our statement marked in red.) So $K$ is not compact.
$\text{*}$ Although not required here (?) but it also follows from finite intersection property of $K$  that the set on right hand side is non-empty.
Is my proof correct? Thanks. 
Edit: As per comments by Mr. Feng Shao, I now realize that there are some mistakes in the proof. But I strongly believe that contrapositive arguments can also prove the statement that I am trying to prove here. Any hint in making this proof work?
 A: Suppose that $E$ has no limit points in $K$. So for every $x \in K$ there is an open ball $B(x, r(x))= \{y\in K: d(x,y) < r(x)\}$ so that $B(x,r(x)) \cap E$ is finite (even at most one point, if you use that definition of limit point)
Then $\{B(x,r_k)\mid x \in K\}$ is an open cover of $K$ so there is a finite subcover
$\{B(x_i, r(x_i)), i=1,\ldots,n\}$.
But then $E$ has at most finitely many points as all $B(x_i, r(x_i)$ only contain finitely many points of $E$..
A: I don't think the proof is right, because the OP didn't use the hypothesis '$E$ is infinite'.
If $E$ has no limit point, then $E$ is closed. For each $x\in E$, beacuse $x$ is not the limit point of $E$, one can find an open set $U_x\in K$ such that $U_x\bigcap E=\{x\}$. Now $\{E^c, U_x|x\in E\}$ is an open covering of $K$. By compactness, there exists $x_1,x_2,\cdots,x_n\in E$ such that
$$K=E^c\bigcup\left(\bigcup_{i=1}^nU_{x_i}\right).$$
It follows
$$E=E\bigcap K=\bigcup_{i=1}^n\left(E\bigcap U_{x_i}\right)=\{x_1,x_2,\cdots,x_n\},$$
which implies that $E$ is a finite set, a contradiction.
This proof doesn't use the metric structure of $K$, so it is valid for general topological spaces.
