I know this claim is false. So what is wrong with the proof? If $A\subseteq B\cup C$, then $A \subseteq B$ or $A\subseteq C$. If $A\subseteq B\cup C$, then $A \subseteq B$ or $A\subseteq C$. A counterexample to this claim is: $A=\{ 2,3,4 \}$, $B=\{1,2,3\}$, $C=\{3,4,5\}$. But I cannot find an error in this proof: Assume $A\subseteq B\cup C$. Let $x\in A$. Then $x\in B\cup C$. So $x\in B$ or $x\in C$. Case 1. $x\in B$. We have $x\in A \rightarrow x\in B$ Thus $A\subseteq B$. Case 2.  $x\in C$. We have $x\in A \rightarrow x\in C$ Thus $A\subseteq C$. Hence, $A\subseteq B$ or $A\subseteq C$.
 A: You have proved that

for every $x$, if $x\in A$, then either $x\in B$ or $x\in C$

Your claim is

(for every $x$, if $x\in A$, then $x\in B$) or (for every $x$, if $x\in A$, then $x\in C$)

With logic symbols, the first formula is

$\forall x\,(x\in A\to ( {x\in B} \lor {x\in C}))$

and the second one is

$(\forall x\,(x\in A\to x\in B))\lor(\forall x\,(x\in A\to x\in C))$

It should be clear that the two are quite different and, indeed, the simple example $A=\{1,2\}$, $B=\{1\}$, $C=\{2\}$ makes the first true and the second false.
If you examine your proof with this example, you see that for $x=1$ we have $x\in B$, but for $x=2$ we have $x\in C$.
A: Draw a Venn diagram where $B$ and $C$ are your normal two circles, and $A$ is a really thin ellipse passing through the middle of the two.
As far as the proof is concerned, you simply forgot that to prove a subset relation you need to show that all elements of one set belong to another. As soon you as you have chosen your $x$ you break into cases, and each time it may be different.
A: From the fact that one element $x$ of $A$ belongs to $B$, you can't deduce that $A\subset B$, since this means that every element of $A$ belongs to $B$..
A: Sometimes it is easier to prove something if you can see that it is true visually.
So try to draw two venn diagrams B and C and color their union. Now let A be another diagram inside the union of B and C, then what can you see from here?
A: If you want to get technical $x \in A \to x\in B$ is ambiguous.  It could mean either $\exists x: x\in A \to x \in B$ or it could mean $\forall x: x\in A \to x \in B$.
In your proof, you are saying "Case 1: $x \in B$ therefore $\exists x: x\in A \to x\in B$".  And that is true.  But the definition of "$A \subset B$" requires the statement "$\forall x: x\in A \to x \in B$".  And that is false.
Consider this:  $2\in A$ .  And $2\in B$.  So the sentence "$2\in A \to 2 \in B$" is true (because $M\to N$ means ($M$ and $N$) or (not $M$)).  So "$\exists x: x\in A \to x \in B$" is true.  But that sure as heck does not mean "$\forall x: x \in A \to x \in B$"! 
Obviously $x = 4$ would be a counter example.
But to say "$A \subset B$" we must have "$\forall x: x\in A \implies x\in B$".  And you just never had that.
Otherwise.... 
we could say:
$\mathbb Z \subset (0,3)$. 
Pf: Let $x = 1$.  $x \in \mathbb Z$ and $x \in (0,3)$.  So $x \in \mathbb Z \to x\in (0,3)$ so $\mathbb Z \subset (0,3)$.
