Projection map being a closed map

Suppose $Z \subset X \times Y$ is closed, and suppose $x_0 \in X \setminus \pi[Z]$. For any $y \in Y, (x_0, y) \notin Z$, and as $Z$ is closed we find a basic open subset $U(y) \times V(y)$ of $X \times Y$ that contains $(x_0, y)$ and misses $Z$. The $V(y)$ cover $Y$, so finitely many of them cover $Y$ by compactness, say $V(y_1),\ldots,V(y_n)$ do. Now define $U = \cap_{i=1}^{n} U(y_i)$, and note that $U$ is an open neighbourhood of $x_0$ that misses $\pi[Z]$: suppose that there is some $(x,y)\in Z$ with $\pi(x,y) = x \in U$. Then $y \in V(y_i)$ for some $i$, and as $x \in U \subseteq U(y_i)$ (as $U$ is the intersection of all $U(y_i)$) we get that $(x,y) \in (U(y_i) \times V(y_i)) \cap Z$ which contradicts how these sets were chosen to be disjoint from $Z$. So $U \cap \pi[Z]=\emptyset$ and $\pi[Z]$ is closed.

To see that the closed projection property implies compactness: suppose $X$ has the closed projection property along $X$, and let $\cal{F}$ be a filter on $X$. Define a space $Y$ that is as a set $X \cup \{\ast\}, \ast \notin X$, where $X$ has the discrete topology and a neighbourhood of $\ast$ is of the form $A \cup \{\ast\}$ with $A \in \cal{F}$. Then $D = \{(x,x): x \in X\}$ is a subset $X \times Y$ and closedness of the projection $p: X \times Y \rightarrow Y$ implies that some point $(x,\ast)$ is in its closure: $D$ cannot be closed in $X \times Y$, because $p[D] = X$ is not closed in $Y$, as $\ast \in \overline{p[D]} = p[\overline{D}]$, by closedness (and continuity) of $p$. This $x$ is an adherence point of the filter, because if $A \in \mathcal{F}$ and $x \in O$ where $O$ is open in $X$, then $O \times (A \cup \{\ast\})$ is basic open in $X \times Y$ and so intersects $D$ in some $(p,p)$, $p \in X$. This $p \in O \cap A \neq \emptyset$, showing that $x$ is an adherence point of $\mathcal{F}$.

There is a standard example for why some hypothesis on $Y$ is necessary: let $X=Y=\mathbb R$, and consider the closed subset $F=\{(x,y)\in \mathbb R\times\mathbb R:xy=1\}\subseteq\mathbb{R}^2$. What is its projection to the first factor?

In fact, one can prove that a space $Y$ is compact iff for all spaces $X$ the projection $X\times Y\to X$ is closed. So while compactness is not necessary (I think...) for the closedness of the projection for one $X$, it is necessary if you want all such projections to be closed.

As for the proof you want in the first bullet point... this is a standard exercise in topology: what have you tried?

This proof is adapted from this lecture note. It is interesting to figure out that this statement is actually a reformulation of The Tube Lemma.

Let $C$ be a closed subset of $X \times Y$, we want to show that $\pi_{1}(C) \subset X$ is closed. To this end, we take any point $x \notin \pi_1(C)$ and show that there exists a neighborhood of $x$ which is disjoint from $\pi_1(C)$.

Since $x \notin \pi_1(C)$, the slice $\{ x \} \times Y$ is disjoint from $C$. Because $Y$ is compact, by The Tube Lemma (replace open with closed and contain with disjoint, respectively), there is a neighborhood $W$ of $x$ such that the whole tube $W \times Y$ is disjoint from $C$. Therefore, $W$ is the neighborhood of $x$ which is disjoint from $\pi_1(C)$, as desired.