Suppose $X$ is a locally compact Hausdorff space and $Y \in \Gamma(X)$. I'll prove that $Y$ is a locally compact Hausdorff space with the subspace topology.
I
Suppose $x,y \in Y$ and $x \not= y$.
- Then $\exists V_i \in \Gamma(X),i \in V_i$ where $i=x,y$ and $V_x \cap V_y = \phi$. This follows from X being Hausdorff.
- For $i=x,y$, put $G_i = Y \cap V_i$. Then $G_i \in \Gamma(Y)$, $i \in G_i$ and $G_x \cap G_y=\phi$.
Hence $Y$ is Hausdorff.
II
Let $x \in Y$. We have:
- $\{x\} \subset X$. $\{x\}$ is $X-$compact.
- By theorem 2.7 of RCA Rudin:
$\;\;\;\;\exists V \in \Gamma(X), \{x\} \subset V \
\subset cl_X(V) \subset Y \
\text{ such that } cl_X(V)\text{ is $X-$compact.}$
$cl_X(V)$ is $Y-$compact as well. Also $V \in \Gamma(Y)$
As $cl_X(V) = cl_X(V) \cap Y$, $cl_X(V)$ is $Y-$closed.
By 2 and 4, $cl_Y(V) \subset cl_X(V)$.
As closed subsets of compact sets are compact,$cl_Y(V)$ is $Y-$compact.
Now we have all the ingredients ready. We have:
- $V \in \Gamma(Y)$ such that $x \in V$.
- $cl_Y(V)$ is $Y-$compact.
Therefore Y is locally compact.
Theorem: $X$ is countably compact, then $X$ is strongly limit compact: every countably infinite subset $A$ has an $\omega$-limit point, i.e. a point $x$ such that for every neighbourhood $U$ of $x$ we have $U \cap A$ is infinite.
Proof: suppose not, then every $x \in X$ has a neighbourhood $O_x$ such that $O_x \cap A$ is finite. Now a nice trick: for each finite subset $F$ of $A$ (and there are countably many finite subsets of $A$) define
$$O(F) = \bigcup\{O_x: O_x \cap A = F\}$$
As every $O_x$ is a subset of one of the $O(F)$ (namely that with $F = O_x \cap A$), and the $O_x$ cover $X$, the $O(F)$ form a countable cover for $X$.
Hence there is a finite subcover $O(F_1), \ldots, O(F_N)$ but then there is some $a_0 \in A \setminus \bigcup_{i=1}^N F_i$ (as the $F_i$ are finite subsets of $A$) and this $a_0$ is not covered by any of the $O(F_i)$ for $i \le N$, and this is a contradiction. So $A$ does have an $\omega$-limit point.
BTW the reverse also holds, as you can see here, from which I also borrowed the above argument.
But now the sequential compactness can be proved: let $(x_n)$ be a sequence in $X$. If $A = \{x_n : n \in \mathbb{N} \}$ is finite, there is a constant (hence convergent) subsequence. So we can assume $A$ is infinite. So $A$ has an $\omega$-limit point $p \in X$, as we saw above.
Let $U_n, n \in \mathbb{N}$ be a countable local base at $p$. Then pick $n_1$ with $x_{n_1} \in U_1 \cap A$. Then having chosen all $x_{n_k} \in (U_1 \cap \ldots U_k) \cap A$, for some $k \ge 1$, where we also have $n_1 < n_2 <\ldots < n_k$, we note that $\cap_{i=1}^{k+1} U_i$ is an open neighbourhood of $p$, so contains infinitely many points of $A$, so in particular we can pick $n_{k+1} > n_k$ such that $x_{n_{k+1}} \in \cap_{i=1}^{k+1} U_i$. Continue this recursion.
It's now standard to check that $x_{n_k} \to p$ is a convergent subsequence of $(x_n)$.
Because I used $\omega$-limit points, there was no need for $T_1$-ness (which you can have with just limit points).
Best Answer
Let $K_n$ be the cofinal sequence of compact subsets and let $p \in X$ and let $U_n$ be a decreasing (WLOG) local base at $p$.
Suppose that for all $n$: $U_n$ is not a subset of $K_n$. Let $x_n \in U_n \setminus K_n$ be a witnessing point of this non-inclusion.
Then $S = \{x_n: n \in \mathbb{N}\} \cup \{p\}$ is a compact subset of $X$ (a sequence together with its limit), and so by assumption on the cofinal sequence, for some $m$, we must have $S \subseteq K_m$. But this cannot be, as $x_m \in S$ and $x_m \notin K_m$. This contradiction shows that the original assumption is false, so that for some $n$, $U_n \subseteq K_n$, which shows that $K_n$ is a compact neighbourhood of $p$.
As this works for all $p$, $X$ is locally compact, as claimed.