Proof $d(A \cup B,C) = \max\{d(A,C),d(B,C)\}$ in any complete metric space

Question

I need to show:

Let $(X,d)$ be a complete metric space. Show that if $A,B,C \in H(X)$ then $d(A \cup B,C) = \max\{d(A,C),d(B,C)\}$ (taken from Fractals Everywhere)

Here $H(X)$ is the set of non-empty compact subsets of $X$. The proposed solutions looks as follows:

$d(A \cup B,C) = \max\{d(x,C):x \in A \cup B\} = \max\{\max\{d(x,C):x \in A\},\max\{d(x,C):x \in B\}\}$

I understand that this is defined for any $d$ so in particular I could apply it to any $d$ operating over $H(X)$. For instance, I want to apply it to the Hausdorff metric. But then, it is not clear why the above equalities hold. Where are we using the fact that the space is complete? Could you give an explanation of this proof?

Answer 1

I will assume that you meant $\min$ instead of $\max$.

Since $d(A\cup B,C)\leqslant d(A,C)$ and $d(A\cup B,C)\leqslant d(B,C)$, then$$d(A\cup B,C)\leqslant\min\bigl\{d(A,B),d(B,C)\bigr\}.\tag1$$Suppose that the inequality in $(1)$ is strict. Take a number $r$ such that$$d(A\cup B,C)<r<\min\bigl\{d(A,B),d(B,C)\bigr\}.\tag2$$Then, there is a $x\in A\cup B$ and a $c\in C$ such that $d(x,c)\leqslant r$. But $x\in A$ or $x\in B$ and therefore one of the numbers $d(A,C)$ and $d(B,C)$ is smaller than or equal to $r$. But this contradicts $(2)$.