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 $\mathbb{K}$.
Let say $\bigcap_{n \in \mathbb{N}} K_n=\emptyset$ . Then $(\bigcap_{n \in \mathbb{N}} K_n)^c=\emptyset^c$. It means $\bigcup_{n \in \mathbb{N}} K_n^c=E$, so the familly $\{K_n^c:n\in \mathbb{N} \}$ is open cover of $E$. Since $E$ is compact we have a finite subcover of the familly. It means there exists a finite subset $S$ of $ \mathbb{N}$ such that $\bigcup_{n \in S} K_n^c=E$. So $(\bigcup_{n \in S} K_n^c)^c=E^c$, it means $\bigcap_{n \in S} K_n=\emptyset$. Buy using decreasing property of the sequence, we have the smallest one of the familly $\{K_n:n\in S\}$. Say it is $K_{n_0}$ and so $K_{n_0}=\bigcap_{n \in S} K_n=\emptyset$. But all the $K_n$'s are non-empty. This is contradiction.
For the third one. $K_{\infty}$ is compact since it is closed and subset of a compact set say $K_n$
Best Answer
We have $$d(a,C)\le d(a,b)+d(b,C)\le d(a,b)+d(B,C) $$ for all $b\in B$, so, taking $\inf_{b\in B}$, we have $d(a,C)\le d(a,B)+d(B,C)$ which is $\le d(A,B)+d(B,C)$ then take $\sup_{a\in A}$, and similarly, we can show that $d(A,c)\le d(A,B)+d(B,C)$ as well.