I remember seeing this proof somewhere (perhaps here, but I don't remember where) that goes something like this.
Suppose $X$ is sequentially compact, and by contradiction suppose $\{U_n\}$ is a countable open cover with no finite subcover. Then for any positive integer $n$, the set $\{U_i : i \le n\}$ is not an open cover, so there exists $x_n \notin \bigcup_{i \le n} U_i$. Hence, we obtain sequence, and by sequential compactness, there exists a subsequence $x_{n_j}$ that converges to $a \in X$. However, $ a \in U_k$ for some positive integer $k$ and by construction, $x_{n_j} \notin U_k$ if $n_j \ge k$. This is a contradiction.
Doesn't this only prove every countable open cover must have a finite subcover?