Set theory. Is the converse true?

**1. The problem statement, all variables and given/known data**

Prove that $\displaystyle \cup_{x \in C} 2^{x} \subseteq 2^{\cup C}$

**2. Relevant equations**

$\displaystyle \cup_{x \in C} 2^{x} = \{ A | \exists x \in C, A \subseteq 2^{x} \}$

$\displaystyle 2^{x}$ is the powerset of $\displaystyle x$. i.e. $\displaystyle 2^{x} = \{ y | y \subseteq x \}$

**3. The attempt at a solution**

Suppose $\displaystyle A \in \cup_{x \in C} 2^{x} $. Then,

$\displaystyle \exists x \in C, A \in 2^{x}$

$\displaystyle \exists x \in C, A \subseteq x$

$\displaystyle A \subseteq ( \cup C )$

$\displaystyle A \in 2^{\cup C}$

Therefore, $\displaystyle A \in \cup_{x \in C} 2^{x} \Rightarrow A \in 2^{\cup C}$

Therefore, $\displaystyle \cup_{x \in C} 2^{x} \subseteq 2^{\cup C}$

But I think there might be something wrong with my proof. Because why can't I start assuming $\displaystyle A \in 2^{\cup C}$ and go to $\displaystyle A \in \cup_{x \in C} 2^{x} $. That means $\displaystyle A \in 2^{\cup C} \Rightarrow A \in \cup_{x \in C} 2^{x} $ and therefore $\displaystyle 2^{\cup C} \subseteq \cup_{x \in C} 2^{x} $ also. That means $\displaystyle \cup_{x \in C} 2^{x} = 2^{\cup C}$

Is there something wrong with this proof?