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.
 A: "since $x \notin$ B and $B \subseteq C, x \notin C$"  That's the mistake. that implication is wrong. Remember the definition of subsets. If $B \subseteq C$, this means that for all elements in $B$, they are also in $C$. Or with symbols $x \in B \rightarrow x\in C$. this statement says nothing about $x$ not being in $B$, so if an element is not in $B$, you can't deduce anything. that's why that implication is wrong. Look, here's a simple example. Let $C = \{w, x, y, z\}$ and $B = \{x, y\}$, so $B$ is a subset of $C$. The element $z$ is not in $B$ but it is in $C$.
A: 
Since $x \not\in B$ and $B \subseteq C$, $x \not\in C$.

Actually, if $x \not\in B$ and $B \subseteq \color{blue}{C}$, both $\color{red}{x \in C}$ and $\color{magenta}{x \not\in C}$ are possible. In a picture:

A: 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 $1\notin B$.
A: Consider:
$C= \{1,2,3,4,5,6,7,8,9,10,11,12\}$, all the integers up to $12$.
And $A = \{2,4,6,8,10,12\}$, all the even numbers up to $12$.
And $B = \{1,4,9\}$, all the perfect squares up to $12$.
We note that $A\subset C$ and $B\subset C$.
That theorem says that if $x\in A$ then $x \in B$.
This is clearly not true.  We have $2,6,8,10,12$ all in $A$ yet none of them is in $B$.
Let's assume $x \in A$.  So $x = 2,4,6,8,10,$ or $12$.
Let's go to the proof.

Suppose $x \not \in B$ 

(okay, so $x \ne 1,4,9$ in particular $x \ne 4$ but $x$ could be $2,6,8,10,12$ still.)

Since $x \in A$ and $A \subset C$, $x \in C$. 

(This follows.  If $x=2,6,8,10$ or $12$ we do have $x \in C$.)

Since $x \not \in B$ and $B\subset C$ then $x \not \in C$.  

Does that actually make sense? $x\ne 1,4,9$ and $1,4,9\in C$ but there are other things in $C$ as well other than those.  We could have $x=2,6,8,10$ or $12$.
There is a concept of $C\setminus B = \{x\in C$ but where $x \not \in B\}$.  As we know that $x\not \in B$ that if $x\in C$ then we must have $x\in C\setminus B$.  But $C\setminus B$ certainly doesn't have to be empty!
So that doesn't follow at all!

But now we have proven both $x ∈ C$ and $x \not \in C$.so we have reached a contradiction. Therefore $x ∈ B$.

Except we haven't proven $x \not \in C$. We have reached no contradiction. And $x$ need not be in $B$.  $x\in C$ so $x$ could be in $C \cap B$ or $x$ could be in $C\setminus B$.  But we have no way of telling.
