Showing the distance between sets is indeed a metric. Let $(X,d)$ be a metric space, and let $A$ be a non-empty subset of $X$. Define the distance between a point and a set by the function
$$d(x,A) = \inf_{z \in A}d(x,z).$$
Prove that for all $x,y \in X$,
$$d(x,A) \leq d(x,y) + d(y,A).$$
Note that it is rather trivial to show that


*

*$d(x,A) \geq 0$ for all $x \in X$ and $A \subseteq X$ (follows directly from non-negativity of a metric);

*$d(x,A) = 0$ iff $x$ is in the closure of $A$; and

*$d(x,A) = d(A,x)$ (by abuse of notation).


By proving the final property of the triangle inequality, we can conclude that the modified distance $d: X \times 2^X \to R$ and its mirrored version $d: 2^X \times X \to R$ forms a sort of pseudo-metric where $d(x,A) = 0$ does not define an equivalence relation.
If we were to modify this function further by first defining $d(x,A)$ to be substituted with $d({x},A)$, we can form a more general pseudo-metric space $(2^X, d:2^X \times 2^X \to R)$ via defining
$$d(A,B) = \inf\{d(a,b) : a \in A, b \in B\}.$$
It should be noted that in this particular metric, $A = B$ means that $A$ and $B$ are not separated. That is, if $A'$ denotes the closure of $A$, then $A \subseteq B'$ and $B \subseteq A'$.
Then we know that (and can verify 4)


*

*$d(A,B) \geq 0$;

*$d(A,B) > 0$ if and only if $A$ and $B$ are separated (i.e., $A$ is not in the closure of $B$, and $B$ is not in the closure of $A$);

*$d(A,B) = d(B,A)$; and

*$d(A,B) \leq d(A,C) + d(C,B)$, for all subsets $C \subseteq X$.


Note that since (2) does not define an equivalence relation on $2^X$, this particular function does not define an actual metric space.
 A: As pointed out in comments, the infimal distance between sets $d(A,B) = \inf\{d(a,b) : a \in A, b \in B\}$, is not a metric. It  fails to satisfy even weaker requirements such as


*

*pseudo-metric (allowed to be zero between different elements)

*quasi-metric (triangle inequality relaxed to $f(A,C)\le K(d(A,B)+d(B,C))$


It's simply unusable as a metric. In contrast, the Hausdorff distance $d_H$ is a metric on the set of nonempty bounded closed subsets, works better. (If one drops "closed", $d_H$ is still a pseudo-metric. If one drops "bounded", it still satisfies the axioms of pseudo-metric but may be infinite sometimes.) 
A: The answer I have found is based on the answer given for this question. Thus, we will generalise this concept to be proven by saying the distance between a point $x$ and a subset $A$ of $X$ forms a bijection with the set distance defined above with $d(x,A) = d(\{x\},A)$. Thus, we will abuse this notation for convenience in the following proof.
Let $a \in A$, $b \in B$, and $c \in C$ be arbitrary points of the subsets $A,B,C$ of $X$. By the inequality
$$d(a,b) \leq d(a,c) + d(c,b),$$
we can take the infimum of the left side with respect to $b \in B$ to get
$$\inf_{b \in B} d(a,b) = d(a,B) \leq d(a,b) \leq d(a,c) + d(c,b).$$
Since our choice of $b$ is independent of $d(a,B)$, we know that for all $b \in B$, $d(a,B) \leq d(a,c) + d(c,b)$. So, take the infimum of the right side with respect to $b \in B$ to get
$$d(a,B) \leq d(a,c) + \inf_{b \in B} d(c,b) = d(a,c) + d(c,B).$$
It is at this point that the original proof is done (but for the special case that $A = \{x\}$ and $C = \{y\}$ where $x,y \in X$), but we will continue to prove the general case. Next, we take the infimum of the left side with respect to $a \in A$ to get
$$\inf_{a \in A} d(a,B) = d(A,B) \leq d(A,B) \leq d(a,c) + d(c,B).$$
Once again, by independence of $a$ and $d(A,B)$, we can take the infimum of the right side with respect to $a \in A$ to get
$$d(A,B) \leq \inf_{a \in A} d(a,c) + d(c,B) = d(c,A) + d(c,B).$$
Edit: Now that we're left with the inequality
$$d(A,B) \leq d(c,A) + d(c,B),$$
we're stuck with the problem of proving we can continue to take infimums while still remaining valid.
