Question concerning the subspace of the Tikhonov Cube of functions with countable support Preparing for a comp, I came across this problem
Let $I$ be the closed unit interval, and $A$ be an uncountable set. $X$ be the set of functions such that $\{a \in A \mid f(a) \neq 0 \}$ is countable ($f: A \to I$ with supp$(f)$ countable).

Show X is sequentially compact but not compact.
Is X separable?
Is X Lindelof?

I think I'm close to being able to show that X is sequentially compact. What I've done:
For fixed $a \in A$, consider a sequence $\{f_n(a)\} \in I$. Since $I$ is compact, $\exists$ convergence subsequence $\{f_{n_k}(a)\}$, letting $f_{n_k}(a)  \to \alpha_a$. We repeat this process $\forall a \in A$ and now we define $f: A \to I$ by $f(a) = \alpha_a$. I was able to fairly easily show $f \in X$ by contradiction, using the fact that the support of each $f_n$ is countable. However, while I'm pretty sure $f$ should be a subsequential limit of $f_n$, I'm not completely sure how to do so.
I THINK that $f$ as I previously constructed it is actually also a cluster point of $f_n$, which would say $X$ is actually countably compact, and since I read that a countably compact Lindelof space is compact, after proving $X$ is not compact I would be able to conclude $X$ is not Lindelof.
However, I have absolutely no idea how to go about showing $X$ is not compact, nor do I have any idea whether or not $X$ is separable.
Any help is appreciated.
 A: $\newcommand{\supp}{\operatorname{supp}}$For $a\in A$ let $I_a$ be a copy of $[0,1]$, so that $X$ is a subspace of $\prod_{a\in A}I_a$. For the proof of sequential compactness, let $S=\bigcup_{n\in\Bbb N}\supp(f_n)$, let $X_S=\prod_{a\in S}I_a$, and let $\pi_S:X\to X_S$ be the obvious projection map. As you’ve already observed, $S$ is countable, so $X_S$ is just the Hilbert cube, a compact metric space. Thus, $X_S$ is sequentially compact, and some subsequence of $\langle\pi_S(f_n):n\in\Bbb N\rangle$ converges to some $f\in X_S$. Now show that if $g\in X$ is defined by $g\upharpoonright S=f$ and $g(a)=0$ for $a\in A\setminus S$, then $\langle f_n:n\in\Bbb N\rangle$ converges to $g$ in $X$. 
The key insight here is that anything to do with a countable subset of $X$ is ‘really’ happening in a subproduct that is a compact metric space. And yes, the same argument shows that $X$ does not have an infinite closed, discrete subset and since $X$ is certainly $T_1$, this shows that $X$ must be countable compact. Thus, showing that $X$ is not compact will also show that $X$ is not Lindelöf, though a direct argument is actually quite easy: consider the open cover $\{U_a:a\in A\}$, where $U_a=\{f\in X:f(a)\ne 1\}$ for each $a\in A$.
To show that $X$ is not separable, let $D\subseteq X$ be countable, and let $S=\bigcup_{f\in D}\operatorname{supp}(f)$. Then $S$ is countable, so there is an $x\in X\setminus S$. Now just define an $f\in X$ such that $x\in\operatorname{supp}(f)$, and find an open nbhd of $f$ disjoint from $D$.
