Elementary Subset Relation Proof I've recently begun learning some elementary set theory, however I have never had to rigorously prove anything before so I've been struggling to construct some elementary proofs, in particular the following exercise:

Prove (A $\subset$ C) $\land$ (B $\subset$ C) $\Leftrightarrow$ (A $\cup$ B) $\subset$ C

I'm constructing it such that I prove the left side implies the right and vice versa. I feel I've proved it left to right but I'm struggling to prove the converse.
From left to right:
Let ($x$ $\in$ A) $\land$ ($y$ $\in$ B)
    $\Rightarrow$ ($x$ $\in$ C) $\land$ ($y$ $\in$ C)
    $\Rightarrow$ ($x,y$ $\in$ C) $\land$ ($x,y$ $\in$ (A $\cup$ B))
    $\Rightarrow$ (A $\cup$ B) $\subset$ C
From right to left:
Let ($x$ $\in$ (A $\cup$ B)) $\land$ ($x$ $\in$ C) $\Rightarrow$ ($x$ $\in$ A) $\lor$ ($x$ $\in$ B)
$\Rightarrow$ (A $\subset$ C) $\lor$ (B $\subset$ C)
You can see I end up only proving that A or B are subsets of C, not strictly that both are subsets. I can see intuitively that the original statement is both true and obvious, but I'm struggling to write it in formal logic so any help as to where I've went wrong would be appreciated.

Actually the left to right is not right. 
To show that $(A \cup B) \subset C$ you need to show that if $x \in A \cup B$, then $x \in C$. That is, you need to work with the same one object $x$, rather than with two objects $x$ and $y$
So in that respect, you actually set it up better in the right to left case, as there you actually are working with just one object.
OK, but for the right to left case. Note that you are trying to show two things: That $A \subset C$, and that $B \subset C$. So, I would think of this as two 'half' proofs: one proof where you assume that $x \in A$, and then show that $x \in C$, and another where you assume that $x \in B$, and then show that $x \in C$. Of course, in both cases you need to use the premise that $(A \cup B) \subset C$
