The probability of an ace from a 5-card hand?

Do you want the probability of $exactly$ one ace? Or just the probability of an ace appearing in a 5-card hand?

If you want exactly one ace, then your answer is correct. $\binom{52}{5}$ is the number of 5-card hands in the deck, and you have 4 choices for which ace to include (hence, $\binom{4}{1}$), and 48 choose 4 choices for the other 4 cards (hence, $\binom{48}{4}$).

If, instead, you want the probability of at least one ace appearing in a 5-card hand, we do things differently. The easiest answer is to find the probability of getting $no$ aces in a 5-card hand.

This probability is $$\frac{\binom{48}{5}}{\binom{52}{5}},$$ for we have 48 choose 5 possible hands with no aces.

Then the solution to the problem - that is, the probability of at least one ace appearing in a 5-card hand - is one minus the complement: $$ 1 - \frac{\binom{48}{5}}{\binom{52}{5}}.$$