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.

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 $X=X\cup Z=Z.$ 
