The sets $X_n$ are closed because (by assumption) the set $X$ has no limit point.
A point $y \in \overline{X_n}\setminus X_n$ would be a limit point of $X_n$, hence a fortiori of $X$.
Your example of $x_n = \frac1n$ in $[0,\,1]$ has a limit point, namely $0$, and therefore the above cannot be applied to it (the $X_n$ are not closed, if you take the intersection of the $\overline{X_n}$, you get a nonempty infinite intersection; the crux is that for any sequence $(x_k)$, the intersection $\bigcap_{k \in \mathbb{N}} \overline{X_k}$ is the set of limit points of $X = \{x_0,\, x_1,\,\ldots\}$).
Your alternative proof is correct, and one of the (many) standard proofs.
Let $X$ be $T_1$. In the proof that "$X$ not countably compact implies $X$ not limit point compact" (the contrapositive) we start with a counterexample to countable compactness:
$\{U_n: n \in \mathbb{N}\}$ a countable open cover of $X$ without a finite subcover.
For each $n$, $\{U_1,\dots,U_n\}$ is not a cover of $X$, so pick $x_n \in X\setminus \cup_{i=1}^n U_i$. In particular:
$$(\ast) \forall n \ge m: x_n \notin U_m$$
Now define $A = \{x_n: n \in \mathbb{N}\}$ then $A$ is infinite.
If not, there is some infinite set of indices and a point $p \in X$ such that
$$\forall n \in B: x_n = p\text{,}$$
because some point $p \in A$ had to be chosen infinitely many times (pidgeon hole principle). But we have a cover and so $p \in U_m$ for some $m$, and then for $n \in B$ with $n > m$ (which surely exists as infinite subsets of $\mathbb{N}$ cannot lie completely below $m$), we would have simultaneously $p=x_n \in U_m$ and $x_n \notin U_m$ (by $(\ast)$), which is clearly absurd. So $A$ is indeed infinite.
Now, if $q \in X$ is any point of $X$, find some $m$ with $q \in U_m$. Then $U_m\cap A \subseteq \{x_1, \ldots x_{m-1}\}$, so is finite, by property $(\ast)$.
Call this finite set $F$. Then for each $f \in F$ such that $f \neq q$ pick a neighbourhood $U_f$ of $q$ such that $f \notin U_f$ by $T_1$-ness.
Then $O:= U_m \cap \bigcap \{U_f: f \in F\setminus \{q\}\}$ is open (as a finite intersection of open sets, contains none of the $f \in F\setminus\{q\}$ so none of the $x_i \in (\{x_1, \ldots x_m\} \cap U_m)\setminus\{q\}$, and is is a subset of $U_m$ so contains none of the $x_n$ with $n > m$. So $O \cap A \subset \{q\}$ which means that $q$ is not a limit point of $A$. So $A$ can have no limit points at all, so $X$ is not limit point compact, as required.
We do need $T_1$ here for the equivalence, otherwise $X = \mathbb{N} \times \{0,1\}$ where $\{0,1\}$ has the indiscrete (trivial) topology and $\mathbb{N}$ the usual discrete one, is an example of a limit point compact space that is not countably compact.
Best Answer
Say $Y=\{a,b\}$. If $S$ is a subset of $\Bbb Z_+\times Y$ and $(n,a)\in S$, then $(n,b)$ is a limit point of $S.$ Conversely, if $(n,b)\in S$, then $(n,a)$ is a limit point of $S.$
There is also the notion of countable compactness: A space is called countably compact if every countable open cover has finite subcover.
Every countably compact space is limit point compact. The converse can fail, as your example demonstrates. This is mainly due to the absence of the $T_1$ property, as every limit point compact $T_1$ space is also countably compact.
Another example of such a space is $X=\Bbb N$ with the topology generated by the sets $A_n=\{k\mid k< n\}$ for all natural $n$. It is limit point compact but not countably compact.