Here's the statement of Theorem 3.6-4 in Erwin Kreyszig's Introductory Functional Analysis With Applications:
Let $H$ be a Hilbert space. Then
(a) If $H$ is separable, then every orthonormal set in $H$ is countable.
(b) If $H$ contains an orthonormal sequence which is total in $H$, then $H$ is separable.
Now my question is, do we require $H$ to be complete? Or, is it not the case that both (a) and (b) above, especially (a), hold even in a general (not necessarily complete) inner product space? At least, the proof given by Kreyszig suggests so.
As noted in my comment above, (a) remains true also for incomplete spaces, because if $H$ is separable, so is the completion $\bar{H}$ (why?) and any orthonormal set in $H$ is also orthonormal in $\\bar{H}$, and hence countable by part (a) for complete spaces.
With the definition of a total set given in the comments, the claim is also true for incomplete spaces, because if the span of a set $M \subset H$ is dense in $H$, it is not hard to see that the $\Bbb{Q}$ span of $M$ is countable and dense in $H$, so that $H$ is separable. (If you use $\Bbb{C}$ as your field, use $\Bbb{Q}+\Bbb{Q}i$ instead of $\Bbb{Q}$).
But if you use the definition that a set $M$ is total if no element $x \in H$ with $x\neq 0$ is orthogonal to all of $M$, then the second claim fails for incomplete $H$, as is implied by GEdgar's answer to this question: A complete orthonormal system contained in a dense sub-space.. He constructs an inner product space such thy every orthonormal set is countable, but which is not separable. If we now take (using Zorn's Lemma) a maximal orthonormal set in that space, it will be total (with the alternative definition) and countable, but the space is not separable.