Q:

Union of a set of collections

I'm given the following definition asked to prove the following theorem:
Definition: Let $X$ be a set and suppose $C$ is a collection of subsets of $X$. Then $\cup \mathbf{C}=\{x \in X : \exists C\in \mathbf{C}(x\in C)\}$
Theorem: Let $\mathbf{C,D}$ be collections of subsets of a set $X$. Prove that $\cup ( \mathbf{C} \cup \mathbf{D}) = (\cup \mathbf{C}) \cup (\cup\mathbf{D})$ 

By my reading of the definition, I run into two problems:
Firstly I think there is a type error since $ ( \mathbf{C} \cup \mathbf{D}) =  \cup\{\mathbf{C}, \mathbf{D}\}$ is not a collection of subsets of $X$ (i.e. it is not a set whose elements are subsets of $X$); instead it is a set whose elements are collections of subsets of $X$.
Second, if we ignore the type error and plug into the definition, we get:
$\cup \{\mathbf{C}, \mathbf{D}\} =\{x\in X:\exists C\in \{\mathbf{C}, \mathbf{D}\}(x\in C)\}=\{x\in X:x\in \mathbf{C} \lor x \in \mathbf{D}\}$, however, since  $\boldsymbol{C},\boldsymbol{D}$ are collections, all their elements are sets. Since $x$ is not a set, $x\notin\boldsymbol{C}\land x\notin\boldsymbol{D}$. Thus $\cup \{\mathbf{C}, \mathbf{D}\}=\emptyset$ 
However this can't be right since the other side of the equality; $(\cup \mathbf{C}) \cup (\cup\mathbf{D})\neq \emptyset$ in general. What am I missing?

A:

Essential are the points:

$a\in\cup b\iff \exists x[a\in x\in b]$
$a\cup b$ is an abbreviation of $\cup\{a,b\}$ (so that $x\in\cup\{a,b\}\iff x\in a\vee x\in b$)

On your first point:
$\mathbf C$ and $\mathbf D$ are collections of subsets of $X$.
Then $\{\mathbf C,\mathbf D\}$ is a set whose elements are collections of subsets of $X$.
Consequently $\cup\{\mathbf C,\mathbf D\}$  is a set whose elements are subsets of $X$, because: $$x\in\cup\{\mathbf C,\mathbf D\}\text{ iff }x\in\mathbf C\text{ or }x\in\mathbf D$$

It must be proved that: $$\cup(\mathbf C\cup\mathbf D=(\cup\mathbf C)\cup(\cup\mathbf D)$$ or equivalently that: $$\cup\cup\{\mathbf C,\mathbf D\}=\cup\{\cup\mathbf C,\cup\mathbf D\}\tag1$$
Equivalent are the following statements:

$x\in\cup\cup\{\mathbf C,\mathbf D\}$
$\exists A\in\cup\{\mathbf C,\mathbf D\}[x\in A]$
$\exists A[[A\in\mathbf C\vee A\in\mathbf D]\wedge x\in A]$
$\exists A[x\in A\in\mathbf D]\vee\exists A[x\in A\in\mathbf D]$
$x\in\cup\mathbf C\vee x\in\cup\mathbf D$
$x\in\cup\{\cup\mathbf C,\cup\mathbf D\}$

This proves $(1)$

