You are correct: that space is not countably compact. Indeed, a space that is countable and countably compact is automatically compact, since every open cover certainly has a countable subcover.
One of the simpler examples of a space that is countably compact but not compact is $\omega_1$, the space of countable ordinal numbers, with the order topology. If for each $\alpha<\omega_1$ we let $U_\alpha=[0,\alpha)$, then $\{U_\alpha:\alpha<\omega_1\}$ is an open cover of $\omega_1$ with no finite subcover. However, every infinite subset of $\omega_1$ has a limit point in $\omega_1$, so $\omega_1$ contains no infinite closed discrete set and is therefore countably compact.
To see this, suppose that $A\subseteq\omega_1$ is infinite. Let $\alpha_0=\min A$, and for each $n\in\omega$ let $\alpha_{n+1}=\min\{\alpha\in A:\alpha>\alpha_n\}$. Then $\langle\alpha_n:n\in\omega\rangle$ is a strictly increasing sequence in $A$. Let $\alpha=\sup_{n\in\omega}\alpha_n$; then $\alpha<\omega_1$, and $\alpha$ is easily seen to be a limit point of $A$.
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
To repeat what was noted in the comments, you’re not showing that $\Bbb R$ is countably compact: you’re showing that it’s Lindelöf. The difference between the two properties lies largely in the placement of the word countable:
Your idea of using the fact that every open set in $\Bbb R$ is a union of open intervals with rational endpoints is a good one. It works: you just have to fill in some details. Let $\mathscr{B}$ be the set of open intervals with rational endpoints, and let $\mathscr{U}$ be any open cover of $\Bbb R$. For each $x\in\Bbb R$ choose a $U_x\in\mathscr{U}$ and a $B_x\in\mathscr{B}$ such that $x\in B_x\subseteq U_x$. (How do you know that this is possible?) Now let $\mathscr{B}_0=\{B_x:x\in\Bbb R\}$.