A topological space The below Theorem is from When are compact and closed equivalent? - Norman Levine
A topological space $( X,\tau)$ will be called a $C - C$ space iff the closed set inn $X$ coincide with the compact sets in $X$.

Theorem 1. Let $\{ (X_i , \tau_i)\}_ {i \in I }$  be a nonempty family of nonempty spaces and let $ ( X,\tau)$ be the product space. if  $ ( X,\tau)$ is $ C - C$, then $\{ (X_i , \tau_i)\}$ is $ C - C $for each $ i \in I $.
Theorem 2. Let $ (X , \tau)$  be a nonempty family of nonempty spaces and let $ ( X \times X, \tau^\star ) $ be the Cartesian product of $ (X , \tau)$  with itself. Then $ ( X \times X, \tau^\star ) $ is $C - C$ iff $ (X , \tau)$ is $ C - C$ and Hausdorff.

Could you give me an explanation of why these are true? I am having difficulty with the proof and the proof in the paper is not clear to me.  
 A: Let $U$ be a closed set in $X_i$ (I'll drop the $\tau$ notation and assume the topology on the space implicitly). Let $p_i\colon X\rightarrow X_i$ be the usual projection map on to the $i$th factor - this map is continuous by definition of the product topology.
Recall that the preimage of a closed set under a continuous map is closed and so $p_i^{-1}(U)$ is closed. By the assumption that $X$ is $C-C$ we know that $p_i^{-1}(U)$ is therefore also compact. Recall also that the image of a compact subset is compact, and so $p_i(p_i^{-1}(U))$ is a compact subset of $X_i$. This set is equal to $U$ and so $U$ is compact.
We can prove this direction a bit faster in the following way. $X$ is $C-C$ and, as $X$ is a closed subset of $X$, we know that $X$ is a compact space. It follows that $p_i(X)=X_i$ is a compact topological space and, as any closed subset of a compact space is compact, we deduce that $U$ must be compact.
Now, suppose that $U$ is a compact subset of $X_i$. The preimage of $p_i^{-1}(U)$ of $U$ is a subset of $X$ which is equal to the product (up to a re-ordering of indices) $$U\times\prod_{j\in {I\setminus\{i\}}}X_i$$ ($U$ has the subspace topology and so is compact as it is a closed subset of a compact space) which is a product of compact spaces and so is compact by Tychanoff's theorem (assuming the axiom of choice). Hence, $p_i^{-1}(U)$ is a compact subset of $X$ and so is closed by the $C-C$ property. Projection maps are open maps and so, as $X\setminus p_i^{-1}(U)$ is open, we must have $p_i(X\setminus p_i^{-1}(U))$ is open as well. It is clear that $$p_i(X\setminus p_i^{-1}(U))=p_i(p_i^{-1}(X_i\setminus U))=X_i\setminus U$$ and so $X_i\setminus U$ is open hence $U$ is closed.
A: I can provide an answer to part (2).
Note first Thomas Andrews's comment above: If a space is C-C, it's compact, so C-C means "compact, with all compact subsets being closed".  Also note that for a Hausdorff space this last condition is automatic, so "C-C and Hausdorff" is the same as "compact and Hausdorff".
Anyway.  Part (2) -- for the backwards implication, note that if $X$ is C-C and Hausdorff, then it's compact and Hausdorff, which means $X\times X$ is compact and Hausdorff, and thus is C-C.
For the forward implication, say $X\times X$ is C-C.  Then it's compact, so $X$ is also compact, being a continuous image of $X\times X$ under the projection map.  Why is it Hausdorff?  Well, in general a space $X$ is Hausdorff if and only if the diagonal $\Delta := \{ (x,x) : x\in X \}$ is closed in $X\times X$.  But $\Delta$ is homeomoprhic to $X$, since the diagonal map $x\mapsto (x,x)$ is continuous, and inverse to either projection map, which is also continuous.  And since $X$ is compact, that means $\Delta$ is compact; since $X\times X$ is C-C, that means $\Delta$ is closed, i.e., $X$ is Hausdorff.  And we already saw earlier that $X$ is compact, so it's compact and Hausdorff, i.e., C-C and Hausdorff.
