Proving this following elementary set theory relation For all sets $A,B,C$ if $A\cap C \subseteq B\cap C$ and $A \cup C \subseteq B\cup C$, then $A\subseteq B$.  
I'm not sure what the second condition is used for because I think I only used the first one:
Let $x \in A\cap C$. Then $x \in A$ and $x \in C$.
Since $A\cap B \subseteq B\cap C$,
$ x \in A$ and $x\in C \implies x\in B$ and $x\in C$.  
Doesn't this automatically show that $x\in A \implies x \in B$?
 A: The second condition is necessary: if say $C=\emptyset$, $A\cap C\subseteq B\cap C$ is trivially true, and we cannot infer $A\subseteq B$.
You need to prove that $x\in A$ implies $x\in B$. As you say this
is the case when $x\in A\cap C$, but what if $x\in A$ but $x\notin C$?
A: No, take for example $A=\{1,2,3\}$, $B=\{1,4,5\}$ and $C=\{1,6\}$.
Then
$$\{1\}=A\cap C \subseteq B\cap C=\{1\}$$
but $A\not\subseteq B$.
As regards the proof of your inclusion, note that
\begin{align*}
A&=(A\cap C)\cup (A\cap C^c)\subseteq (B\cap C)\cup ((B\cup C)\cap C^c)
\\&\subseteq (B\cap C)\cup (B\cap C^c)\cup (C\cap C^c)\\
&=(B\cap C)\cup (B\cap C^c)=B.
\end{align*}
A: It is not sufficient that $x \in A \cap C\implies A \in B \cap C$ alone implies $A \subset B$. For example, let $B$ and $C$ be the empty set, and let $A$ be the set of natural numbers.
However, if the other condition is true, then the result follows, for this reason:
If $x \in A$, then either $x \in C$ or $x \notin C$.
If $x \in C$, then $x \in A \cap C \subset B \cap C$, so $x \in B$.
If $x \notin C$, then $x \in A \cup C$ [which is true anyway]  $\subset B \cup C$, so $x \in B$ or $x \in C$ must be true, but we know that the latter proposition is not true, hence the former must be true, so $x \in B$.
Either way, $x \in B$, so that $A \subset B$. 
