Hausdorff distance and union of sets Let $X$ be a metric space; $A_1$, $A_2$, $B_1$, $B_2$ be non-empty subsets in $X$. Let $d(\cdot,\cdot)$ be the Hausdorff distance between sets in $X$. Then
$$
d (A_1 \cup A_2 , B_1 \cup B_2) \leq \max \{ d(A_1,B_1), d(A_2,B_2)\}.
$$
 A: For a subset $A \subset X$ and a real $\epsilon>0$, let $\cup_{\epsilon}A$ denotes $\{x \in X \mid d(x,A)< \epsilon\}$.
Let $r =\max (d(A_1,B_1),d(A_2,B_2))$. Then $A_i \subset \cup_rB_i$ and $B_i \subset \cup_r A_i$. Therefore, $$\cup_r(B_1 \cup B_2) = \cup_rB_1 \cup \cup_r B_2 \supset A_1 \cup A_2$$ and in the same way $$\cup_r(A_1 \cup A_2) \supset B_1 \cup B_2.$$ Thus, $d(A_1 \cup A_2, B_1 \cup B_2) \leq r$.
A: This is somehow related to my question How do max and union commute in Hausdorff measure?. A solution for your problem appears in Barnsley's book on Superfractals, theorem 1.12.15, page 66. Let me follow his approach and borrow some notation from my linked question. I will also assume the algebraic definition of Hausdorff distance, unlike the previous answerer, but both definitions are equivalent.
You want to prove that $d_H(A \cup B,C \cup D) \le \max\{d_H(A,C),d_H(B,D)\}$. On the linked answer you will find a proof of $d(B \cup C, A) = \max\{d(B,A),d(C,A)\}$. From here it is direct that
$$d(C \cup D, A \cup B) = \max\{d(C,A \cup B),d(D,A \cup B)\}.\qquad (1)$$ 
Then one establishes:
$$\begin{align*}d(C,A \cup B) &= \max_{c \in C} \min_{x \in A \cup B} d(c,x)\\
&= \max_{c \in C} \min\{\min_{a \in A} d(c,a),\min_{b \in B} d(c,b)\}\\
&\le \min\{ \max_{c \in C} \min_{a \in A} d(c,a),\max_{c \in C} \min_{b \in B}d(c,b)\}\\
&= \min\{d(C,A),d(B,C)\}.\end{align*}$$
As a consequence $d(C,A \cup B) \le d(C,A)$ and $d(D,A \cup B) \le d(D,B)$. Now, we come back to $(1)$ and subsitute, so we get $d(C \cup D,A \cup B) \le \max\{d(C,A),d(D,B)\}$ and $d(A \cup B,C \cup D) \le \max\{d(A,C),d(B,D)\}$
Finally, 
$$\begin{align*}d_H(A \cup B, C \cup D) &= \max\{d(C \cup D, A \cup B),d(A \cup B,C \cup D)\}\\
&\le \max\{\max\{d(C,A),d(D,B)\},\max\{d(A,C),d(B,D)\}\}\\
&\le \max\{d(C,A),d(D,B),d(A,C),d(B,D)\}\\
&= \max\{\max\{d(C,A),d(A,C)\},\max\{d(B,D),d(D,B)\}\\
&= \max\{d_H(A,C),d_H(B,D)\}.\end{align*}$$
Note the author is working in $H(X)$ the set of non-empty compact subsets of your base space but the proof is the same changing $\max$ for $\sup$ and $\min$ for $\inf$. I welcome any edits that can improve the layout of the equations above. 
