Why is this proof invalid? I don't understand why this theorem is false.

Suppose that $A \subseteq C$, $B \subseteq C$, and $x \in A$. Then $x \in B$.

Invalid Proof:

Suppose that $x \notin B$. Since $x \in A$ and $A \subseteq C$, $x \in C$. Since $x \notin B$ and $B \subset C$, $x \notin C$. But now we have proven both $x \in C$ and $x \notin C$, so we have reached a contradiction. Therefore $x \in B$.

I'm thinking that $$\forall x(x \in B \implies x \in C) = \forall x(x \notin B \vee x \in C)$$
It's true that $x \notin B$, so this statement is true, and can't be used to prove through contradiction that $x \in B$. But I'm not completely sure, so any clarification would help here. Thanks.

Here's a counter model . . .
Let $A=\{1\}, B=\{2\}, C=\{1,2\}$. Then $A\subseteq C$ and $B\subseteq C$ with $1\in A$ but $1\notin B$.
