Confusion on wrong proof on conditions for equivalence of compactness and countable compactness. I was exploring Topology by Hocking, and came across a problem which requires a proof given 'countable compactness' [1], 'completely separable' [2], and Hausdorff, but I couldn't see why Hausdorffness is necessary.


*

*Countable compactness means that every infinite subset has a limit
point

*Completely separable means there is a countable basis
The question is "Prove that a subset of a completely separable $T_2$ space is compact if and only if it is countably compact."
The only if part is easy, since every compact set is countably compact, as was proven on page 19.
In a prior exercise(exercise 1-13, page 21), proven was a space is countably compact if and only if given a countable collection of closed sets of which finite selection of has nonempty intersection, then the intersection of all the closed sets is nonempty. If I take contrapositives, this means that if given a countable collection of closed sets, if the total intersection is empty, then there must be a finite number of closed sets whose intersection is empty.
Then, to prove the 'if' part...
Let us pick a subset $X$, and a cover $\{U_a\}$ of $X$, then there is a countable subcover of $\{U_a\}$ by Lindelof's theorem proved on page 66. Then, we can order them by the naturals. The sets $V_n=X \cap U_n$ is open in the subspace topology for X, and $X-V_n$ is a closed set in the subspace topology, since $X-(X-V_n)=V_n$ for $V_n \subset X$. We note that $\bigcup^{\infty}_{n=1} V_n=X$, for otherwise, there is a point $p$ in $X$ which no $V_n$, and thus $U_n$ covers. Then, $X-\bigcup^{\infty}_{n=1} V_n=\emptyset$, but by De Morgan's Law in generalized form, $X-\bigcup^{\infty}_{n=1} V_n=\bigcap^{\infty}_{n=1} X-V_n$, which likewise was found to be empty. But, then it follows by the above contrapositive of the theorem proven in exercise 1-13, there are a finite subset of this collection of closed sets which has empty intersection, that is, for simplicity, let take them to be labeled by the first $n$ naturals. Then, $\bigcap^n_{i=1} X-V_i=\emptyset$, but then similarly, by De Morgan's Law in set form, $\bigcap^n_{i=1} X-V_i=X-\bigcup^n_{i=1} V_i=\emptyset$, or to say, that $X=\bigcup^n_{i=1} V_i$, or that $\bigcup^n_{i=1} U_i$ covers $X$, or to say that it is a finite subcover.
But, I take notice that all I used was countable compactness, complete separability, but not Hausdorffness.
I looked at the possibility that the exercise 1-13 is wrong, but it doesn't appear so: take the closed sets, and consider a sequence formed by interescting the prior elements with $C_n$, if there is a point at which there is no point in the sequence which is a subset of that before it, then it automatically has nonempty intersection, otherwise pick points in $C_1\bigcap...C_n$ not in $C_1\bigcap...C_m$ for some $m>n$, and in this process, obtain an infinite sequence of points which has a limit point, and since every closed set contains its limit points, is contained in all the closed sets or the entire intersection, or to say it is nonempty. For the reverse, every countable sequence of points, were it to not have a limit point, has no subset have a limit point either, and contains all its limit points, and so, the sets obtained by removing the naturals $1$ to $n$ is a sequence of closed sets which has empty intersection, a contradiction.
I know Lindelof's theorem isn't wrong, so clearly something is wrong with my proof, and it does feel a bit roundabout, but I have reviewed what I could that mentions De Morgan's Laws but I still don't see what is missing.
To be clear, I am not wondering about what the proof is but what the error in my proof, or framing of a property, is.
 A: The Hausdorff part is strictly speaking unnecessary (but maybe your text requires compact spaces to be Hausdorff; this is common practice). 
The essential argument is: completely separable implies Lindelöf and in all spaces Lindelöf + countably compact (in the "real" meaning: that a countable cover has a finite subcover !) is equivalent to compactness.
But "every infinite set has a limit point" is equivalent to "every countable cover has a finite subcover" for $T_1$ spaces. (There are non-$T_1$ counterexamples, like $X=\Bbb N \times \{0,1\}$, the two point space in the indiscrete topology; a "fat doubling"of $\Bbb N$ which obeys the limit point condition, but not the countable cover one.)
So only $T_1$-ness (implied by Hausdorff) is needed, really.
The example I gave above also shows 1.13 is false for limit point compactness: let $C_n = X \setminus (\{0,1,\ldots,n\} \times \{0,1\})$ which is a sequence of closed decreasing sets with empty intersection, while if $A \subseteq X$ is any non-empty subset (say $(n,i) \in A$ for some $n \in \Bbb N, i \in \{0,1\}$, then $(n,1-i)$ is a limit point of $A$ as its minimal open neighbourhood is $\{(n,i), (n,1-i)\}$ which intersects $A$ in another point than itself, namely $(n,i)$.
The exercise becomes true if we replace the limit point condition by 

for all infinite $A \subseteq X$ there is a point $p \in X$ such that every open neighbourhood of $p$ contains infinitely many points of $A$ ($p$ is called an $\omega$-limit point in some texts).

and is then equivalent (in all spaces) to "every countable open cover of $X$ has a finite subcover", so "real" countable compactness. 
See this older answer for more details.
