prove there is a unique set $X$ such that every set $Y$, $Y∪X = Y$ For this proof. It seems obvious that $X=∅$ such that for every set $Y$,  $Y∪X = Y$ since the $Y∪X$ is just Y. How should I go about this?
Let there be sets $X,Z$
Since $Y∪X = Y$ then,
{$x| x ∈ Y ∨ x ∈ X$} = {$x| x ∈ Y$}
Since $Y∪Z = Y$ then,
{$x| x ∈ Y ∨ x ∈ Z$} = {$x| x ∈ Y$}
Then equate the two 
{$x| x ∈ Y ∨ x ∈ Z$} = 
{$x| x ∈ Y ∨ x ∈ X$}
Therefore conlude Z = X and prove that X is a unique set.
 A: First, check that $X=\emptyset$ works. Now, suppose there is another possibility $X\neq\emptyset$ that satisfies the requirement. Then, take $Y=\emptyset$, and note that
$$
Y\cup X\neq Y
$$
because the RHS is empty whereas the LHS equals $X$ and therefore has at least 1 element.
A: As Kim Jong Un said, the first is show that $\emptyset$ satisies your requirements.
Now, let if $X$ has the property, then, taking $Y=\emptyset$, $X\subseteq \emptyset\cup X=\emptyset$. Thus $X\subseteq \emptyset$ and done.
A: You need one more step after your last equation: If $\{x :x\in Y\lor x\in Z\}=Y=\{x:x\in Y\lor x\in X\}$ holds for every $Y$, then it holds when $Y=X$: $$X=\{x:x\in X\lor x\in X\}=\{x: x\in X\lor x\in Z\}.$$ And it holds when $Y=Z$: $$Z=\{x:x\in Z\lor x\in Z\}=\{x:x\in Z\lor x\in X\}.$$ Comparing the RHS of these, we have $X=Z.$........ More succintly we can obtain $$X=X\cup Z$$ by applying $\forall Y\;(Y=Y\cup Z)$ to the case $Y=X,$ and we can obtain  $$Z=X\cup Z$$ by applying $\forall Y\;(Y=X\cup Y)$ to the case $Y=Z.$ .... Thus $X=X\cup Z=Z.$ 
A: If there is such a set  X then:
$X \cup \emptyset =\emptyset  $
But $X \cup \emptyset = X $
So if such a set exists, it must be the empty set.
The empty set is such a set, so such a set does exist and it is uniquely the empty set.
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2nd argument.
Let S be any set that isn't the empty set.  Let $x \in S $.  Let $Y $ be any set that does not contain the $x $.
Then $x \in S \cup Y $ so $S \cup Y \ne Y$.
So S doesn't have the property.  So only the empty set has the property.
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Argument 3:
Let X be such a step.  Then $X^c \cup X = X^c $.  So $X^c $ contains no elements of $X $.  But $X^c \cup X $ contains all elements of X.  
So if such a set exist it has no elements.  It must be the empty set.
