Proving closed subsets of compact sets (in a metric space) are compact. Let $X$ be a metric space and let $K\subset X$ be a compact set in $X$ and let $E\subset K$ be closed subset of $K$. 
I want to prove that $E$ is compact. I have asked this question before also but this time I am trying to prove it differently than the last time. 
Proof: I'll prove: $E$ is not compact $\implies E $ is not closed relative to $K$. Then contrapositive of it proves the statement that I'm trying to prove. 
There exists an open cover $(U_\alpha)$ of $E$ without a finite subcover. 
Here $(U_\alpha)=\{U_\beta \subset K: \beta \in \text{index set and } U_\beta \text{ is open relative to $K.$}\}$ such that $E\subset (U_\alpha)$.
If $E$ is closed relative to $K$ then clearly the set $K - E$ is open rel. to $K$. And therefore, $(K-E)\cup (U_\alpha)$ forms an open cover of $K$ and therefore, $K\subset (K-E)\cup (U_\alpha)$. And since $K$ is compact $\exists$ finitely many indices $\alpha_1,\alpha_2,\cdots,\alpha_n$ such that $K\subset (K-E)\cup (U_{\alpha_i}) $, where $i=1,2,\cdots, n$. $\tag 1$
It follows that $E\subset K\subset (K-E)\cup (U_{\alpha_i}) \implies E\subset (U_{\alpha_i}) $ and therefore open cover $(U_\alpha)$ has a finite subcover, which is a contradiction and therefore $E$ is not closed.
Note: Since $K\subset K \subset X$, then it can be proven that $K$ is compact in $K$ if and only if $K$ is compact in $X$. Therefore, open covers of $K$ even if they come from $K$ will have a finite subcover. And therefore, there exist $\alpha_1,\cdots,\alpha_n$ in $(1)$ above.
Is my proof correct? Thanks.
 A: One small point needs clarification. When we say $K$ is compact in $X$ we mean that any cover of $K$ by open sets in $X$ has  finite subcover. But you have used the fact that any cover of $K$ by open sets in $K$ has  finite subcover. This is not wrong but you have to say why this is true.
A: Seems correct to me except for the fact that Kavi Rama Murthy pointed out but I also think that you are kind of proving the statements directly, since you are doing the contrapositive of the Lemma and then you're supposing that $E$ is closed in $K$.
What would have changed if you simply started with $E$ being closed in $K$ and by "absurd" not compact? The exact same words would have been used.
An other way to prove the statement would have been :
Let $E \subset K$ be a closed set, where $K$ is compact. Let $\{U_\alpha\}$ be an open cover of $E$. Then $\{U_\alpha\} \cup \{E^c\}$, where $E^c$ is the complement of $E$ w.r.t. to $X$, covers $K$. Since $K$ is compact, we can extract a finite subcover $\{E^c, U_{\alpha_1}, U_{\alpha_2}, \ldots, U_{\alpha_n}\}$ from $\{U_\alpha\} \cup \{E^c\}$. Since $E \cap E^c = \varnothing$, we have that $\{ U_{\alpha_1}, U_{\alpha_2}, \cdots, U_{\alpha_n}\}$ is a subcover of $\{U_\alpha\}$
