First of all, "countably" is redundant, because there are only countably many sentences in the first place. With that out of the way, you are correct. Assume there are only finitely many unprovable statements (up to equivalence modulo ZFC) $\sigma_1 \ldots \sigma_n$, and that ZFC is consistent. Then take $\text{ZFC}^\star$ some complete, consistent extension of ZFC. Since there are only finitely many $\sigma_i$, we can effectively axiomatize $\text{ZFC}^\star$ by taking the axioms of ZFC together with one of $\sigma_i$ or $\
eg \sigma_i$ for each $i$ (depending on whether $\text{ZFC}^\star \models \sigma_i$), contradicting the first incompleteness theorem.