Theorem 3.6-4 in Erwin Kreyszig's Introductory Functional Analysis With Applications 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. 
 A: 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. 
