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.

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.
