You need to show that $\neg A \lor \neg B$ implies a contradiction. So, showing that $\neg A$ by itself implies a contradiction is not enough, since $\neg A$ is not implied by $\neg A \lor \neg B$.
So, you need to show that each of $\neg A$ and $\neg B$ lead to a contradiction.
Then you would have $\neg A \to \bot$ as well as $\neg B \to \bot$, which means that you have $(\neg A \to \bot) \land (\neg B \to \bot)$, which is equivalent to $(\neg A \lor \neg B) \to \bot$, which is what you need to show.
This makes sense from the following perspective as well: Typically, with a goal like $A \land B$, you would show two things: that $A$ is implied by premise $P$, and that $B$ is also implied by premise $P$. Now, to show that $A$ is implied by $P$, you can do a proof by contradiction: show that $\neg A$ leads to a contradiction. But clearly you then still need to prove $B$ as well ... and for that you could do its own proof by contradiction. So again, you need both.
Let me also point out that approaching the problem from the latter perspective (proving $A$ and $B$ separately) is really just the best way to approach this, because you may find that, for example, proving $A$ might work well using a proof by contradiction, but for proving $B$, a direct approach might work a lot better. So, this way you keep your options open. But if you assume the negation of the conclusion $Q$, you basically lock yourself into doing a proof by contradiction for both.