Prove that there is a unique $A\in\mathscr{P}(U)$ such that for every $B\in\mathscr{P}(U), A\cup B = B$ $U$ can be any set.
For the existence element of this proof, I have $A = \varnothing$
But it's for the uniqueness element of this proof where I am having trouble.  So far I have:
$\forall(C\in\mathscr{P}(U))(\forall(B\in\mathscr{P}(U))(C\cup B=B)\rightarrow C\in\mathscr{P}(U) = A\in\mathscr{P}(U))$
Assume $C\in\mathscr{P}(U)$ is arbitrary and assume $\forall(B\in\mathscr{P}(U))(C\cup B=B)$ then:
What I have left to prove is: 
$C\in\mathscr{P}(U) = A\in\mathscr{P}(U) $ 
which is where I am having trouble.  I can say that $C = \varnothing$ which would then equal $A$ but it is not certain that $C = \varnothing$. 
 A: For uniqueness: The claim has to hold in particular for $B = \varnothing$. So if $A \neq \varnothing$, what would happen?
A: For the uniqueness if $A$ and $A'$ are two sets verifying the hypothesis then we have
$$A\cup A'=A'=A'\cup A=A$$
so $A=A'$ and obviously this set is $\emptyset$
A: Here is an attempt at a slightly more constructive answer, where $\;\varnothing\;$ is not pulled like a rabbit out of a magician's hat.
First a point of notation: noticing that $\;X\in\mathscr{P}(U)\;$ is equivalent to $\;X \subseteq U\;$, I will implicitly assume in this answer that $\;A\;$ and $\;B\;$ range over subsets of $\;U\;$.  This gets rid of a lot of occurrences of $\;\in\mathscr{P}(U)\;$.
Using this notation we are to prove that there is a unique $\;A\;$ such that
$$
(0) \;\;\; \langle \forall B :: A \cup B = B \rangle
$$
Apart from simplifying this to
$$
(1) \;\;\; \langle \forall B :: A \subseteq B \rangle
$$
using set theory, the only way to make progress seems to be to choose a specific $\;B\;$.  And there three values of $\;B\;$ immediately suggest themselves: $\;\varnothing\;$, $\;A\;$, or $\;U\;$.  Since the last two only lead to tautologies, we choose $\;\varnothing\;$.  (There are also combinations of these, like $\;U \setminus A\;$ which also turns out to work but in a more complex way.  However, the simplest option already works, as we will see.)
Therefore we calculate for any $\;A\;$
\begin{align}
& \langle \forall B :: A \subseteq B \rangle \\
\Rightarrow & \;\;\;\;\;\text{"choose $\;B := \varnothing\;$ -- as discussed above"} \\
& A \subseteq \varnothing \\
\equiv & \;\;\;\;\;\text{"set theory: simplify using $\;\varnothing \subseteq X\;$ for any $\;X\;$"} \\
& A = \varnothing \\
\end{align}
And the same law that was used in the last step above can now immediately be used for the other direction: assuming $\;A = \varnothing\;$ we directly prove $(1)$ by
\begin{align}
& \langle \forall B :: A \subseteq B \rangle \\
\equiv & \;\;\;\;\;\text{"assumption"} \\
& \langle \forall B :: \varnothing \subseteq B \rangle \\
\equiv & \;\;\;\;\;\text{"set theory: $\;\varnothing \subseteq X\;$ for any $\;X\;$"} \\
& \textrm{true} \\
\end{align}
Putting these together we see that $(0)$ is equivalent to
$$
A = \varnothing
$$
Therefore the unique $\;A\;$ satisfying $(0)$ is $\;\varnothing\;$.
A: If $U$ is empty then this is trivial.  If $U$ is not empty then suppose some such $A\ne \varnothing$ exists.  Let $x\in A$.  There is at least one subset of $U$ not containing $x$, since $\varnothing$ is such a set.  For any subset $B$ of $U$ such that $x\not\in B$, we have $A\cup B\ne B$.
