Show that if $A \subseteq B $ then $ A \cup C \subseteq B \cup C$ for all sets, A, B, C, if $A \subseteq B  $ then $  A \cup C \subseteq B \cup C$
So if every element of A is also an element of B then the elements of A $\cdot$ C are identical to the elements of B $\cdot$ C
Suppuse $x\in A$
Then, since $A\subseteq B$ we have $x \in B$. 
Hence, $x\in A \cup C $ and $x \in B \cup C$
Therefore, $A \cup C \subseteq B \cup C$
 A: You've only shown that $A \subseteq B \cup C$. The warning sign is your phrase "suppose $x \in A$" - from that point forward, you're only paying attention to members of $A$. But the goal, $A \cup C \subseteq B \cup C$, says that every member of $A \cup C$ regardless of whether it is in $A$ or in $C$ is also a member of $B \cup C$.
So you need another case - something that starts with "suppose $x \in C$". If you can show that under that assumption instead you also have $x \in B\cup C$, then you're done.
A: Suppose $x\in A\cup C$ then either $x\in A$ or $x\in C$. If $x\in A$ then $x\in B$ as $A\subseteq B$ and thus $x\in B\cup C$. If $x\in C$ then $x\in B\cup C$
A: Your proof is missing a case.
We are given that $x \in A \to x \in B$
We want to show that $x \in A \ \text{or} \ C \to x \in B \ \text{or} \ C$
So there are two cases: $x \in A$, $x \in C$
Case 1: $x \in A$
$x \in A \to x \in B \to x \in B \ \text{or} \ C$
Case 2: $x \in C$
$x \in C \to x \in B \ \text{or} \ C$
Another way is to think of three cases:  $x \in A$ but not $C$, $x \in C$ but not $A$ and $x \in A$ and $C$
Left as an exercise
