How do you prove $X\cup (X\cap Y) = X$? I;m not sure how to go about proving this. I just started learning about it and would appreciate some help.
 A: When dealing with equalities about sets, a good technique is to show that the left side of the equation is always a subset of the right side, and that the right side is also a subset of the left. Thus, the sets are equal. In your case, you would want to prove that 
$$
X \cup (X \cap Y) \subseteq X\ \ \ (1)
$$
and
$$
X \subseteq X \cup (X \cap Y)\ \ \  (2).
$$
Now, for $(1)$, if $x \in X \cup (X \cap Y)$ then either $x$ belongs in $X$ or in $(X \cap Y)$ (the latter meaning that $x$ is both in $X$ and $Y$). So either way, you have that $x \in X$. For $(2)$, if $x \in X$ then $x$ belongs in $X \cup (X \cap Y)$ since $x \in X$ and you just need to prove that it belongs to just one of $X$ or $(X \cap Y)$. 
Since both $(1)$ and $(2)$ hold, we have that every element of the left side also belongs to the right, and every element of the right also belongs to the left. The only way this can be true is when $ X = X \cup (X \cap Y)$.
A: Claim 1:  $X\cap Y \subset X$.
Proof:  If $x \in X\cap Y$ then $x \in X$ and $x \in Y$.  So $x \in X$.
Claim 2:  If $A \subset X$ then $X = X\cup A$.
Pf: a: if $x \in X$ then $x\in X$... so $x \in X$ or $X\in A$ is true.  So $x \in X\cup A$.
and $X\subset $X\cup A$.
If $x \in X\cup A$ then either $x \in X$ or $x \in A$.  If $x \in A$ then $x \in X$ because $A\subset X$.  On the other hand if $x \in X$ then $x\in X$.  So either way $x \in X$.
so $X\cup A \subset X$.
So $X= X\cup A$.
So claim 1 and 2 together prove your result:
....
Or directly.  If $x \in X$ then $x\in X$ so $x \in X$ or $x \in X\cap Y$ is true.  So $X \subset X\cup (X\cap Y)$.
If $x \in X\cup (X\cap Y)$ then either $x \in X$ or $x \in (X\cap Y)$.  If $x \in X$ ... then $x \in X$.  If however $x \in X\cap Y$ then $x \in X$ and $x \in Y$.  So $x \in X$. Either way $x \in X$ so $X\cup(X\cap Y) \subset X$.
....
