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.

This sort of argument is best written out in plain language, with minimal use of symbols.
Suppose $x\in A$. That is, $x$ is an element of $A$. Then certainly $x$ is either an element of $A$ or an element of $B$ (or both). Thus $x$ is an element of $A\cup B$, ie. $x\in A\cup B$.
Since we have shown that any element of $A$ is an element of $A\cup B$, $A\subseteq A\cup B$.
