Let $E = \mathbb{R}^N$ for some $N \in \mathbb{N}$ and $d(x,y)=||x-y||_2$ the euclidean distance. We note $K(E)$ the set of all compact sets of $E$, and we'll use the Hausdorff distance on $K(E)$ defined as $D_H(A,B)=\max(\delta(A,B), \delta(B,A))$ where $\delta(A,B)=\sup_{x\in A}\operatorname{dist}(x,B)$.
Now, I am asked to do this:
Verify that a subset $F \subset K(E)$ is bounded for the distance $D_H$ if and only if $\bigcup_{A \in F}A$ is a bounded subset of E.
So here's what I tried, but I am very unsure if it's acceptable as a proof.
Let $F \subset K(E)$ be bounded for $D_H$. Then for a fixed set $A \in F$, there exists $M \in \mathbb{R}$ such that for all $B \in F$ we have $D_H(A,B) \leq M$. Then as $B' = \bigcup_{A' \in F}A' \in F$ we have $D_H(A,B') \leq M$, thus $B'$ is bounded.
Now, suppose that $\bigcup_{A' \in F}A'$ is bounded, as $F \subset \bigcup_{A' \in F}A'$ we immediately have that $F$ is bounded.
Is this enough?
Neither part works. Problem 1: $$B' = \bigcup_{A' \in F}A' \in F$$ There is no reason for that union to be an element of $F$. The family $F$ is not assumed to be closed under unions.
Problem 2: $F \subset \bigcup_{A' \in F}A'$. This doesn't even make sense; the union on the right is a set in $\bigcup_{A' \in F}A'$, but $F$ is a family of sets, it's not any subset of $\mathbb{R}^n$. Consider carefully the distinction between "set of sets" and "union of sets".
Another way to see the proof is unlikely to be correct: you haven't used the definition of Hausdorff distance at all.
A correct proof could go like this.
Let $F \subset K(E)$ be bounded for $D_H$. Then there exist $A \in F$ and $M \in \mathbb{R}$ such that for all $B \in F$ we have $D_H(A,B) \leq M$. The set $A$ is compact, hence it is contained in some ball $\mathbb{B}(c, R)$. If $b$ is a point of some set $B\in F$, then by the definition of the Hausdorff distance we have $\operatorname{dist}(b, \mathbb{B}(c, R))\le M$. By the triangle inequality this implies $d(b, c)\le M+R$. Thus, $B\subset \mathbb{B}(c, M+R)$. Since this is true for every $B\in F$, the union of all such $B$ is also contained in $\mathbb{B}(c, M+R)$ and is therefore bounded.
Conversely, suppose the set $\bigcup_{A \in F}A$ is bounded. Let $M$ be its diameter. By the definition of diameter, we have $d(p, q)\in M$ for all $p, q\in \bigcup_{A \in F}A$. Pick any two sets $A,B\in F$. For any points $a \in A$ and $b\in B$ we have $d(a, b)\le M$ since both $a$ and $b$ are in the aforementioned union. By the definition of Hausdorff distance, $d_H(A, B)\le M$. It follows that $F$ is bounded.