TITLE: How can I prove this statement about subsets?
QUESTION [7 upvotes]: Let $A$, $B$, $C$ be sets. Prove that if $A \subseteq C$ and $B \subseteq C$ then $A \cup B \subseteq C$.

This is an exercise in mathematical logic.
My attempt to progress forward: This statement can be written as$$
(A \subseteq C) \land (B \subseteq C) → A \cup B \subseteq C\\
(x \in A → x \in C) \land (x \in B → x \in C) → A \cup B \subseteq C
$$
But I am not even sure that is how I am supposed to do it so I am a bit stuck. Can anyone explain to me how to get through this proof? Thanks in advance.

REPLY [2 votes]: I haven't noticed one in the in the answers, so here's a proof by contradiction. 
The statement we wish to prove is:
$$(\forall x)(x \in A \cup B \implies x \in C) \tag{1} $$
It's negation is:
$$ (\exists x)(x \in {A} \cup {B} \wedge \neg(x \in C) ) \tag{2}   $$
$$ x \in A \cup B \implies (x \in A \vee x \in B)$$
$$  x \in A \wedge A \subseteq C \implies x \in C \tag{3}  $$
$$ or$$
$$  x \in B \wedge B \subseteq C \implies x \in C \tag{4} $$
Since both $(3)$ and $(4)$ are in contradiction with $(2)$, it must be false. Therefore our starting premise $(1)$ is correct.