Consider the statement: "Prove that $X \cup (Y \setminus X) \subseteq X \cup Y$" (I believe the two unions are actually equal, but for the sake of this question, that is not relevant).
Converting this into a first order statement that I want to prove (if this is incorrect, please let me know and I will change it for clarity):
$$\forall a,X,Y\big(a \in X \cup (Y\setminus X) \rightarrow a \in X \cup Y\big)$$
Note the following:
$$a\in X \cup (Y \setminus X) \iff a \in X \lor (a \in Y \land a \notin X)$$
$$\neg(a \in Y \setminus X) \iff a \notin Y \lor a \in X$$
Now, to prove the above universal statement, I would start from all cases where the antecedent evaluates to true. I will break down all of these cases (one of which has 3 subcases). The cases or subcases that have "contradictions" will be highlighted in $\color{red}{\text{red}}$.
Case 1: In English..."$a$ is in $X$, but $a$ is not in $Y \setminus X$"
$a \in X \text{ is True }, (a \notin Y \lor a \in X) \text{ is True}$
$\color{red}{\text{Subcase 1A}}$: $a \in X \text{ is True}, a \notin Y \text{ is True}, a \notin X \text { is True}$
$\text{Subcase 1B}$: $a \in X \text{ is True}, a \in Y \text{ is True}, a \in X \text{ is True}$
$\text {Subcase 1C}$: $a \in X \text{ is True},a \notin Y \text{ is True},a \in X \text{ is True}$
Case 2: In English... "$a$ is not in $X$, but $a$ is in $Y \setminus X$"
$a \notin X \text{ is True }, (a \in Y \land a \notin X) \text{ is True}$
$\text{Subcase 2A}$: $a \notin X \text{ is True }, a \in Y \text{ is True}, a \notin X \text{ is True}$.
$\color{red}{\text{Case 3}}$: In English... "$a$ is in $X$, and $a$ is in $Y \setminus X$"
$a \in X \text{ is True}, (a \in Y \land a \notin X) \text{ is True}$
$\color{red}{\text{Subcase 3A}}$: $a \in X \text{ is True}, a \in Y \text{ is True}, a \notin X \text{ is True}$
So my question basically breaks down to the following: What do I do with these contradictions? Do I simply "ignore" them and only work with the "non-contradictions"?
If I am supposed to ignore them, why exactly is that so? I don't really understand what a contradiction IS when it comes to how one makes logical arguments. What would happen if all cases were purely contradictions? Would the implication (from which these cases derive) simply be vacuously true? Cheers~
Long story short: Nope, you don't have to worry about those contradictions.
$X$ and $Y$ are free variables, so should not be quantified. $X\cup(Y\smallsetminus X)\subseteq X\cup Y$ equates to the claim:
$$\forall a~\big(a\in X\cup(Y\smallsetminus X)\to a\in X\cup Y\big)$$
Somewhat. The antecedent is true when the arbitrary entity $a$ is in $X$ or it is in $Y\smallsetminus X$. You only have to consider these two cases, which may or may not be disjoint. If both cases each entails the conclusion, and those two cases cover all possibilities, then the conclusion is derivable.
$$\dfrac{P\vdash R\quad Q\vdash R\quad\vdash P\vee Q}{\vdash R}$$
Note: Indeed, when they are disjoint, their conjunction will be contradictory. However, the Principle of Explosion applies: "from a contradiction, anything follows" (ex falso quodlibet). So, you do not have to worry about the conjunction entailing a contradiction.
(Another way to view it: the intersection $X\cap(Y\smallsetminus X)$ is an empty set, which is a subset of any set. That is $a\in X\cap(Y\smallsetminus X)\to a\in (X\cup Y)$ holds vacuously.)
So...
Taking an arbitrary element $a$, and assuming that $a\in X\cup(Y\smallsetminus X)$ we have two exhaustive cases:
Case 1: $a\in X$
Case 2: $a\in Y\smallsetminus X$
So $a\in X\cup (Y\smallsetminus X)$ does entail $a\in X\cup Y$. Since that $a$ was arbitrary, we conclude $\forall a~(a\in X\cup(Y\smallsetminus X)\to a\in X\cup Y)$, which proves that which was to be proven by definition of subset$_{eq}$ .
$$\therefore~ ~X\cup(Y\smallsetminus X)\subseteq X\cup Y$$