Proof that $X \subseteq Y$ if and only if $X \cup Y = Y$ This exercise (2.1.49) was taken from An Infinite Descent into Pure Mathematics.
Question
Let $X, Y$ be sets. Prove that $X \subseteq Y$ if and only if $X \cup Y = Y$.
Proof
Towards a contradiction, assume $X \not \subseteq Y$ if and only if $X \cup Y = Y$.
For $X \not \subseteq Y$, then $X$ must be non-empty and contain a set of elements, $N$ such that $N$ is disjoint with $Y$. Therefore, assume $X = \{x : x \in Y \vee x \in N\}$.
For $X \cup Y = Y$, $N = \{\}$. This is a contradiction with the assumption that $N$ is inhabited. Therefore, the assumption $X \subseteq Y$ if and only if $X \cup Y = Y$ cannot be true. We have shown that $X \subseteq Y$ if and only if $X \cup Y = Y$. $\Box$
Is this proof rigorous enough, and are there any improvements that can be made?
 A: To prove a statement of the form $A \iff B$, you need to show that both implications $A \implies B$ and $B \implies A$ hold. I will sketch both directions but leave a few details for you to fill in.

First let's prove $X \subseteq Y \implies X \cup Y = Y$.

To do this, we assume that $X \subseteq Y$, and we need to prove that $X \cup Y = Y$. To prove equality of two sets, we need to prove containment in both directions: namely $X \cup Y \subseteq Y$ and $Y \subseteq X \cup Y$.
To prove the first containment $X \cup Y \subseteq Y$, start with the assumption $X \subseteq Y$. This implies that $X \cup Y \subseteq Y \cup Y$ (why?), and of course $Y \cup Y = Y$, so we conclude that $X\cup Y \subseteq Y$ as desired.
The second containment $Y \subseteq X \cup Y$ is always true and doesn't even require the assumption that $X \subseteq Y$ (why?)

Now let's prove $X \cup Y = Y \implies X \subseteq Y$.

So, assume that $X \cup Y = Y$. The containment $X \subseteq X \cup Y$ always holds (why?). By the assumption, the RHS of this containment equals $Y$. Thus $X \subseteq Y$ as desired.
A: Let $x$ be an arbritary element in $X$. Then,
$$x \in X \cup Y.$$
Also,
$$ X \cup Y = Y \implies x \in Y.$$
Since $\forall x \in X,\ x \in Y$, $X \subseteq Y.\tag1$
Let $y$ be an arbritary element in  $X \cup Y$. Then,
$$X \cup Y = Y \implies y \in Y$$
Since $\forall y \in X \cup Y,\ y \in Y,$
$$\begin{aligned} X \cup Y \subseteq Y \\
\implies X \subseteq Y \end{aligned}\tag2$$
From $(1)$ and $(2)$,
$$X \subseteq Y \iff X \cup Y = Y.\ \blacksquare$$
A: Below a visually organized proof in two columns, one for each direction ( of the biconditional that has to be proved) 

A: Let $X,Y$ be sets. 
$(\Rightarrow)$ Suppose $X\subseteq Y$. First we show that $X\cup Y\subseteq Y$. Let $x\in X\cup Y$. 
Case 1. $x\in X$. If $x\in X$, then $x\in Y$ since $X\subseteq Y$. Thus $X\cup Y\subseteq Y$. 
Case 2. $x\in Y$. If $x\in Y$ then we are done. Again $X\cup Y\subseteq Y$.
So we conclude that $X\cup Y\subseteq Y$. 
Now we show that $Y\subseteq X\cup Y$. Let $x\in Y$. Since $x\in Y$, $x\in X\cup Y$. So we conclude that $Y\subseteq X\cup Y$. We've shown that  $X\cup Y\subseteq Y$ and $Y\subseteq X\cup Y$, so $Y = X\cup Y$.
$(\Leftarrow)$ Suppose $X\cup Y = Y$. We need to show that $X\subseteq Y$. Let $x\in X$. Since $x\in X$, $x\in X\cup Y$. Since $X\cup Y = Y$, we have $x\in Y$. So $X\subseteq Y$ which is what we needed to show. 
Thus $X\subseteq Y$ iff $X\cup Y = Y$. 
QED
