How to continue this proof about Powersets? I need to prove the following statement:
Prove that if $a \in A$, then $P(A - \{a\}) \cap \{C \cup \{a\}\ |\ C \in P(A - \{a\})\} = \emptyset$, where $P(A)$ is the powerset of set $A$.
My try:
Suppose $X \in P(A - \{a\}) \cap \{C \cup \{a\}\ |\ C \in P(A - \{a\})\} = \emptyset$
By the definition of $\cap$, we know that $X \in P(A - \{a\})$ and $X \in \{C \cup \{a\}\ |\ C \in P(A - \{a\})\}$.
By the definition powerset we also know that $X \subset A - {a}$ and $X \subset \{C \cup \{a\}\ |\ C \in P(A - \{a\})\}$
By negation law, we know that either $a \in X$ or $a \notin X$.
Case 1: Suppose $a \in X$.
By the definition of $\subseteq$, $a \in A - \{a\}$. Then, $a \in A$ and $a \notin \{a\}$, which is a contraditiction because $a \in \{a\}$.
Case 2: Suppose $a \notin X$.

This is where i lost myself. I can see that $a$ must be in the set $\{C \cup \{a\}\ |\ C \in P(A - \{a\})\}$, but I don't get how this leads to a contradiction.

If $X\in\mathcal{P}(A\setminus\{a\})$, then $a\notin X$. If $X\in\bigl\{C \cup \{a\}\,|\,C \in\mathcal{P}(A \setminus \{a\})\bigr\}$, then $a\in X$. Therefore, no set can belong to both $\mathcal{P}(A)$ and to $\bigl\{C \cup \{a\}\,|\,C \in\mathcal{P}(A \setminus \{a\})\bigr\}$. In other words, $\mathcal{P}(A)\cap\bigl\{C \cup \{a\}\,|\,C \in\mathcal{P}(A \setminus \{a\})\bigr\}=\emptyset$.
