Prove or disprove $A$ compact/closed $\implies$ $\mathcal{P}(A)$ compact/closed 
For every $A \subset \mathbb R^3$ we define $\mathcal{P}(A)\subset \mathbb R^2$ by
  $$ \mathcal{P}(A) := \{ (x,y) \mid \exists_{z \in \mathbb R}:(x,y,z) \in A \} \,. $$
  Prove or disprove that $A$ closed $\implies$ $\mathcal{P}(A)$ closed and $A$ compact $\implies$ $\mathcal{P}(A)$ compact.

This $\mathcal{P}(A)$ seems to me the projection on the $xy$-plane, so intuitively both make sense to me.
My try
For the first one, to prove something is closed seems easiest to do with sequences, so for all sequences $(x^{(n)})$ in $A$ with  $x^{(n)} \to x$ we have that $x \in A$. But I do not know how to progress further in a useful direction. For the second one I don't even know where to start.
Update
For the first one I tried to argue that for any sequence going to $x \in A$, a projection of that sequence on the $xy$-plane will also go to the projection of $x$, so $x \in \mathcal{P}(A)$, so hence $\mathcal{P}(A)$ must be closed. 
 A: Let $A=\{(x,0,z): xz=1\},$ which is closed in $R^3.$ Then $P(A)=\{(x,0):x\ne 0\},$ which is not closed in $R^2.$
Observe that $P$ is Lipschitz-continuous, as $\|P(u)-P(v)\|=\|P(u-v)\|\leq \|u-v\|.$ The continuous image of a compact space is compact. So if $A\subset R^3$ is compact then $P(A)$ is compact.
A: Hints:


*

*The fact that you are unable to prove this might suggest that the statement is actually untrue.  Try and work out where your proof is falling apart and use that to construct a closed set $A\subset \mathbb R^3$ such that $P(A)$ is not closed.

*This is not as difficult as you're making it out to be.  What definition of compactness are you using?
A: Hint:
The first one is not true.
For the second one try this:
I assume you know Heine Borel's theorem about compact sets.
Heine Borel says $A$ is bounded. So for all components of $(x_1,x_2,x_3)\in A $ you can find an $M>0$: $|x_i|<M$. That means $|x_i|^2<M^2$. Now we have:
\begin{align} ||(x_1,x_2)||^2=|x_1|^2+|x_2|^2<2M^2\end{align}
Can you finish now?
Heine Borel says $A$ is also closed. Choose a convergent sequence $x^n \rightarrow x$ in $\mathcal{P}(A)$. There is a sequence $x^n_3$ with $(x_1^n,x_2^n,x_3^n)\in A$. You know $x_3^n$ is boundend. By Bolzano Weierstrass theorem for sequence we have now there is $x^{n_k}_3 \rightarrow x_3$. What can you say about the original sequence in $\mathcal{P}(A)$ now?
Combine them both and by Heine Borel you have $\mathcal{P}(A)$ is compact.
