Proving a set is a subset of a union with itself and another set Let A and B be sets. Show that
$$\space A \subseteq (A \cup B)$$
My proof is as follows, but I don't feel confident that what I've done is correct. Any input is much appreciated.
$$x \in A \cup B \equiv x \in A \wedge x \in B \equiv x \in A$$
Therefore, A is a subset of the union of A and B.

You're going in the wrong direction. Don't assume $x\in A\cup B$, assume $x\in A$: you want to show that each element in $A$ is also an element of $A\cup B$ (not vice versa). Also, $x\in A\cup B\iff x\in A\textrm{ OR (not and) }x\in B$.
Let $x\in A$, and note that $x\in A\cup B\implies x\in A$ or $x\in B$. But, as $x\in A$, $x\in A\cup B$. Therefore, each element of $A$ is an element of $A\cup B$, so that $A\subseteq A\cup B$.
However, you should note that your proof is the correct proof of $A\cap B\subseteq A$ if you replace your $A\cup B$ with $A\cap B$.
