Characterization of countably compact space. I have some troubles with the next problem.
Let $D$ be a countable set and $p\notin D$. Consider the space $Y=D\cup\{p\}$ with the next topology: Every $d\in D$ is isolated and if $U\subseteq Y$ such that $p\in U$, then $U$ is open if and only if $D\setminus U$ is finite. Let $X$ a $T_2$ space. Prove that $X$ is countably compact if and only  if the projection $\Pi_{Y}:X\times Y\rightarrow Y$ is a closed function.
Proof
$\Rightarrow)$
Clearly, $Y$ is compact. We need to prove the next two lemmas.
Lemma 1 If $f:X\rightarrow Y$ is a closed mapping defined onf a Hausdorff space $X$ and all fibers $f^{-1}[\{y\}]$ are countably compact, then for every countably compact subspace $Z\subseteq Y$ the inverse image $f^{-1}[Z]$ is countably compact.
Proof of Lemma 1
Let $U$ be an open cover of $f^{-1}[Z]$ composed by open sets of $X$. We define $\mathcal{C}=\left\{\displaystyle\bigcup U_0:U_0\subseteq U \ \text{and} \ |U_0|<\omega\right\}$. Note that for all $U\in\mathcal{C}$, the set $Y\setminus f[X\setminus U]$ is open. With this, the set $\mathcal{V}=\left\{ Y\setminus f[X\setminus U]:U\in \mathcal{C}\right\}$ is an open cover of $Z$. Why? Let $y\in Z$, then, $f^{-1}[\{y\}]\subseteq f^{-1}[Z]\subseteq \displaystyle\bigcup U$. Because $f^{-1}[\{y\}]$ is countably compact, then there exist $U_0\subseteq U$ finite such that $f^{-1}[\{y\}]\subseteq \displaystyle\bigcup U_0$. Let Be $W=\displaystyle\bigcup U_0\in \mathcal{C}$. Then, $f^{-1}[\{y\}]\subseteq W$ and finaly, $y\in Y\setminus f[X\setminus W]$. Therefore, $\mathcal{V}$ covers $Z$.
Because $Z$ is countably compact, then there exist $\mathcal{V_0}\subseteq \mathcal{V}$ finite such that $Z\subseteq \displaystyle\bigcup \mathcal{V_0}$, i.e., there exist $\mathcal{C}_0\subseteq \mathcal{C}$ finite sucha that $Z\subseteq\displaystyle\bigcup_{U\in \mathcal{C}_0} Y\setminus f[X\setminus U]$. Therefore, $f^{-1}[Z]\subseteq \displaystyle\bigcup \mathcal{C}_0$. Then, if $K\in \mathcal{C}_0$, then there exist $\mathcal{U}_u\subseteq U$ finite such that $K=\displaystyle\bigcup \mathcal{U}_u$. Then, $\displaystyle\bigcup_{K\in\mathcal{C}_0}\mathcal{U}_K$ is the finite subcover that we want.
Lemma 2 If $X$ and $Y$ are Hausdorff spaces such that $X$ is countably compact and $Y$ is compact then $X\times Y$ is countably compact.
Consider the projection $\Pi_{X}:X\times Y\rightarrow X$. This function is closed because $Y$ es compact. Let $x\in X$. Then, $\Pi^{-1}_{X}[\{x\}]=\{x\}\times Y$ and clearly $\{x\}\times Y$ is homeomorphic to $Y$. Then, because $Y$ is compact, $\Pi^{-1}_{X}[\{x\}]$ is compact and then, countably compact. By the Lemma 1, if we consider $X$ (we know that $X$ is countably compact) then $\Pi^{-1}_{X}[X]=X\times Y$ is countably compact.
Finally, the implication. Let $F\subseteq X\times Y$ a closed set. We want to prove that the projection $\Pi_{Y}: X\times Y\rightarrow Y$ is closed. Because $X\times Y $ is countably compact, then $F$ is countably compact. Because the projection is continuous, $\Pi_{Y}[F]\subseteq Y$ is countably compact. Because $Y$ is $T_2$ and first countable, then  $\Pi_{Y}[F]$ is closed.
But, with the another implication $\Leftarrow)$, I don't have idea. Some hint? I thought to use the characterization of countably compact via accumulation points, but, really I don't know how.
 A: $Y$ is just a convergent sequence with limit (so $\omega+1$ (in the order topology)or $\alpha(D(\aleph_0))$ (the Aleksandrov-compactification of a discrete countably infinite space). 
I'll use that a space $X$ is countably compact iff every infinite subset has an $\omega$-accumulation point (see here e.g.)
Then the argument is pretty straightforward:
So consider the projection. $\pi: X \times (\omega+1) \to \omega+1$.
Let $C$ be closed in the product and let $x \in \overline{\pi[C]}$; we want to see that $x \in \pi[C]$. If $x = n$ for some $n$, $\{x\}$ is a neighbourhood of $x$ that should intersect $\pi[C]$ which means that $x \in \pi[C]$. So we only need to consider $x = \omega$. Then either $x \in \pi[C]$ already (and we are done) or $\omega \notin \pi[C]$ and $\pi[C]$ is an infinite subset of $\omega$ and so we have infinitely many $n$ with $n \in \pi[C]$, so that we have an infinite set $N \subseteq \omega$ with $(x_n, n) \in C$ for all $n \in N$ and some $x_n \in X$. The set $\{x_n: n \in N\}$ has an $\omega$-accumulation point $q \in X$ (from countable compactness of $X$) and then $(q,\omega) \in \overline{C} = C$, so that indeed $\omega \in \pi[C]$, as required. So when $X$ is countably compact, $\pi$ is a closed map.
Suppose for the reverse that $\pi$ as above is a closed map and suppose that $A = \{a_n: n \in \omega\}$ is a countable subset of $X$ without an $\omega$-accumulation point. Then $C = \{(a_n, n): n \in \omega\}$ is closed in $X \times (\omega+1)$ (as is easily checked: no point $(x,n)$ can be in the closure (this uses $T_1$-ness of $X$) and neither can a point like $(x,\omega)$ be) and this implies that $\omega = \pi[C]$ would be closed in $\omega+1$, which is false. So if $\pi$ is closed, $X$ is countably compact.
