Let be $(X,d)$ a metric space, $K_1 , K_2 \subset X$ compact and closed, respectively. Consider $$A = \{ d(x,y) \in \mathbb{R}_+ : x \in K_1 , y \in K_2 \}$$
Is $A$ compact?
Remark: If $X$ is not complete, then $A$ is not compact. For example, if $X = \mathbb{R}-\{0\}$. Take $K_2 = (0,1]$ and $K_1 = \{1\}$. Then $A= [0,1)$ is not compact.
But if $X$ is complete. It is true?
For $A$ to be compact, it is necessary that $A$ is bounded. Hence, if $K_2$ is an unbounded closed set, then $A$ cannot possibly be compact. This gives you as much counterexamples as you’d like.
Completeness has nothing to do with all of this.
It is sufficient that $K_2$ is compact. Here’s a proof.
Let $a_n = d(x_n,y_n)$ be a sequence in $A$ convergent to some $a$. The sequence $(x_n)$ has a convergent subsequence $(x_{n_k})$ to some point $x\in K_1$. The sequence $(y_{n_k})$ has a convergent subsequence $(y_{n_{k_t}})$ to some point $y \in K_2$. The sequence $(x_{n_{k_t}})$ is a subsequence of $( x_{n_k})$, so it is convergent to the same point $x$. Then $(a_{n_{k_t}})$ is a subsequence of $(a_n)$ converging to $d(x,y)$. Hence $(a_n)$ converges to $d(x,y)$ and therefore $a = d(x,y) \in A$. Therefore, $A$ is closed.
As $K_1$ and $K_2$ are bounded, so is $K_1 \cup K_2$, so there is $M>0$ such that for each $x\in K_1$ and each $y\in K_2$, $d(x,y)\le M$. So $A$ is bounded above by $M$. We conclude that $A$ is compact.
P.S. A simpler way to prove compactness is to note that $A = f(K_1 \times K_2)$, where $f: K_1 \times K_2 \to \mathbb R$ is given by $f(x,y) = d(x,y)$. We know that $f$ is continuous (in fact Lipschitz), and that $K_1 \times K_2$ is compact since each $K_i$ is compact. Hence $A$ is compact being a continuous image of a compact set (though, to prove that the product of two compact sets is compact, you do an argument similar to the one I did in the answer, so in some sense this isn’t a “simpler way”).