You need the $T_1$ axiom for one direction, though. Otherwise let $\{0,1\}$ have the indiscrete topology, and define $X = \mathbb{N} \times \{0,1\}$. There every non-empty set $A$ has a limit point (if $(x,0) \in A$, then $(x,1)$ is a limit point of $A$ and the same with $0$ and $1$ interchanged), but $\{n\} \times \{0,1\}, n \in \mathbb{N}$ is a countable cover without a finite subcover.
Or you could use another (less standard) notion of limit point, where $x$ is a limit point of $A$ iff every neighbourhood $O$ of $x$ has $O \cap A$ infinite. This is usually called an ($\omega$)-accumulation point, or a strong limit point, or some such name.
The proof for $T_1$ spaces can be found here.
For the strong limit point version: suppose $X$ is countably compact (in the cover sense) and let $A$ be an infinite subset of $X$ that has no strong (!) limit point. We can just assume $A$ is countable or we pass to an infinite subset of it. Then by assumption on $A$, for every $x \in X$ we have an open set $O_x$ such that $O_x \cap A$ is finite. To construct a countable cover without a finite subcover define for every finite $F \subseteq A$ the set $O_F = \bigcup \{O_x: O_x \cap A = F \}$. There are at most countably many finite subsets of a countable set, so $\{O_F: F \subseteq A \text{ finite } \}$ is a countable cover of $X$ (every $O_x$ is in exactly one of the $O_F$'s and the $O_x$ were a cover).
Suppose $O_{F_1},\ldots,O_{F_n}$ were a finite subcover of this open cover. There is some $a_0 \in A \setminus \cup_{i=1}^n F_i$, and this $a_0$ cannot be covered by these finitely many sets $O_{F_i}$, by construction. This contradicts the countable compactness, so $A$ must have a strong limit point.
For the reverse direction, suppose every infinite subset has a strong limit point (i.e. $\omega$-accumulation point) and suppose that $\{U_n: n \in \mathbb{N} \}$ is a countable open cover of $X$ without a finite subcover. So pick $x_1 \in U_1$ and note that $U_1$ cannot cover $X$ so we pick $x_2 \notin V_1 := U_1$. But $x_2$ is covered by some $U_{n_2}, n_2 > n_1$, and we define $V_2 = \bigcup_{i=1}^{n_2} U_i$. Then $V_2 \neq X$ (or we'd have a finite subcover), so pick $x_3 \notin V_2$. Again, $x_3 \in U_{n_3}$ for some $n_3 > n_2$, define $V_3 = \bigcup_{i=1}^{n_3} U_i$ etc. By recursion we get a set $x_n$ of all distinct points and increasing open sets $V_n$ such that $x_n \in V_n$ but $x_n \notin V_k$ for $k < n$. As the $n_k$ keep increasing, $\cup_n V_n = \cup_n U_n = X$.
Now $x$ is a strong limit point of $\{x_n: n \in \mathbb{N}\}$, which must exist. Then $x \in V_m$ for some $m$, but this $V_m$ does not contain any of the $x_k$ with $k > m$ so cannot intersect this set in an infinite set, contradiction. So $X$ is countably compact.
So for this stronger notion of limit point the statement holds. Note that in a $T_1$ space, $x$ is a strong limit point of $A$ iff it is a limit point of $A$, so if you know that, you can apply the above proof for the $T_1$ case as well.
(*) In general space $Y$ is compact if and only if every family of closed sets that has finite intersection property has a non-empty intersection.
For a proof of that see here.
Choose some $K\in\mathcal H$.
Let $\mathcal H_K$ denote the collection $\{K\cap H\mid H\in\mathcal H\}$ of subsets of $K$ and now focus on compact Hausdorff space $K$.
Prove that $\mathcal H_K$ has finite intersection property and is a collection of sets closed in compact set $K$.
Applying (*) it can be concluded that $\bigcap\mathcal H_K\neq\varnothing$ and consequently $\bigcap\mathcal H\neq\varnothing$.
Best Answer
This uses the fact that $S_i$ is closed. You know $S_i$ intersects every neighborhood of $x_0$, and so $x_0$ is in the closure of $S_i$. Since $S_i$ is closed, this tells you that actually $x_0\in S_i$.