Claim: A topological space $\,X\,$ is compact iff it has the Finite Intersection Property (=FIP):
Proof: (1) Suppose $\,X\,$ is compact and let $\,\{V_i\}\,$ be a family of closed subsets s.t. $\,\displaystyle{\bigcap_{i}V_i=\emptyset}\,$. Putting now $\,A_i:=X-V_i\,$ , we get that $\,\{A_i\}\,$ is a family of open subsets , and
$$\bigcap_{i}V_i=\emptyset\Longrightarrow \;X=X-\emptyset=X-\left(\bigcap_iV_i\right)=\bigcup_i\left(X-V_i\right)=\bigcup_iA_i\Longrightarrow\;\{A_i\}$$
is an open cover of $\,X\,$ and thus there exists a finite subcover of it:
$$X=\bigcup_{i\in I\,,\,|I|<\aleph_0}A_i=\bigcup_{i\in I\,,\,|I|<\aleph_0}(X-V_i)=X-\left(\bigcap_{i\in I\,,\,|I|<\aleph_0}V_i\right)\Longrightarrow \bigcap_{i\in I\,,\,|I|<\aleph_0}V_i=\emptyset\Longrightarrow$$
The family $\,\{V_i\}\,$ has the FIP.
(2) Suppose now that every family of closed subsets of $\,X\,$ hast the FIP, and let $\,\{A_i\}\,$ be an open cover of it. Put $\,U_i:=X-A_i\,$ , so $\,U_i\, $ is closed for every $\,i\,$:
$$\bigcap_iU_i=\bigcap_i(X-A_i)=X-\bigcup_i A_i=X-X=\emptyset$$
By assumption, there exists a finite set $\,J\,$ s.t. $\,\displaystyle{\bigcap_{i\in J}U_i=\emptyset}\,$ , but then
$$X=X-\emptyset=X-\bigcap_{i\in J}U_i=\bigcup_{i\in J}X-U_i)=\bigcup_{i\in J}A_i\Longrightarrow$$
$\,\{A_i\}_{i\in J}$ is a finite subcover for $\,X\,$ and thus it is compact....QED.
Please be sure you can follow the above and justify all steps. Check where we used Morgan Rules, for example, and note that we used the contrapositive of the FIP's definition...
Suppose $A$ is compact. Then let $x$ be in $X \setminus A$. We must find a neighbourhood of $x$ that is disjoint from $A$ to show closedness of $A$.
For every $a \in A$ let $U_a = B(a, \frac{d(a,x)}{2})$ and $V_a =B(x, \frac{d(a,x)}{2})$. Then $U_a$ and $V_a$ are disjoint open neighbourhoods of $a$ and $x$ respectively (disjoint follows from the triangle inequality, check this).
Then the $U_a ,a \in A$ together form a cover of $A$, and now we use compactness of $A$ to get finitely many neighborhoods $U_{a_1}, \ldots, U_{a_n}$, that also cover $A$. But then $V_{a_1} \cap \ldots V_{a_n}$ is an open neighbourhood of $x$ that misses $A$ (since a point in $A$ cannot be in all of $V_{a_1}, \ldots, V_{a_n}$ because it belongs to some $U_{a_i}$ which is disjoint with $V_{a_i}$.).
So $A$ is closed.
An alternative with sequences: suppose $x \in \overline{A}$. Then being a point in the closure we find a sequence in $A$, $(a_n)$ that converges to $x$. The sequence $(a_n)$ from $A$ has a convergent subsequence in $A$ by compactness of $A$, so there is some $a \in A$ and some subsequence $a_{n_k} \rightarrow a$. But also $a_{n_k} \rightarrow x$, and as limits of sequences in metric spaces are unique: $a =x$, but then $x \in A$ as required: $\overline{A} = A$.
Best Answer
It is probably easier to appeal to the Heine-Borel Theorem
Step (1): Show that $D$ is a metric on $\mathbb{K}$.
Now that $(\mathbb{K},D)$ is a metric space, by the Heine-Borel theorem, it suffices to show that it is complete and totally bounded.
Step (2): Let $A_i$ be a Cauchy sequence in $\mathbb{K}$, show that it converges. (In fact, you can show that as long as $M$ is a complete metric space, then so is $\mathbb{K}$ with the metric you just wrote down.)
Step (3): Show that for every $\epsilon > 0$ there exists a finite open cover of $\mathbb{K}$ by $\epsilon$-balls. (In fact, as long as $M$ is totally bounded, so will $\mathbb{K}$.)
Step (1) is obvious, so I won't give a hint.
For Step (2), let $A_i$ be your Cauchy sequence. Consider the set $A$ of points $a$ such that for any $\epsilon > 0$, $B(a,\epsilon)$ intersects all but finitely many $A_i$.
For Step (3), start with a finite cover of $M$ by $\epsilon$ balls, let $S$ be the set of the center points of those balls. Consider the power-set $P(S)$ (the set of all subsets of $S$) as a set of points in $\mathbb{K}$.