Both of the other answers here do not follow <PERSON>'s approach directly, though they are still roughly correct. Here is an answer which exactly follows his directions, and shows most of the details. Based on the fact that you're talking about QM, I assume you are using complex Hilbert spaces, though there's no substantive difference for real Hilbert spaces.
In my answer, I've assumed some familiarity with real analysis, but not functional analysis (since this is a basic construction from functional analysis). If you don't know what a Cauchy sequence in a metric space is, you'll want to familiarize yourself with that before starting on part 2. Also, this answer really has no physics; it's all functional analysis (the study of infinite dimensional vector spaces).
1. Why does $V$ have the structure of an inner product space?
He has defined $V$ as the subset of $\times_\alpha {\cal H}\alpha$ where all but finitely many $\Psi\alpha$ vanish. That is to say, a generic element of $V$ is a choice of $\Psi_\alpha$ for every alpha, where the only nonzero elements are $\Psi_{\alpha_1}, \Psi_{\alpha_2}, \ldots, \Psi_{\alpha_n}$ for some positive integer $n$ and some choice of indices $\alpha_1, \ldots, \alpha_n$. Note that different elements $\Psi$ and $\Psi'$ will in general correspond to different positive integers $n$, $n'$ and different indices $\alpha_1, \ldots, \alpha_n$ and $\alpha'1, \ldots, \alpha'{n'}$. For the rest of this section, I will take $\Psi$ and $\Psi'$ to be two generic elements of $V$ given by the above formulas.
We want to put an inner product on $V$. 364
Normally, for finite direct sums, we'd want to do something like $\langle \Psi | \Psi' \rangle = \displaystyle \sum_\alpha \langle \Psi_\alpha | \Psi'\alpha \rangle{\alpha}$, where $\langle \cdot | \cdot \rangle_\alpha $ is the inner product on $\mathcal H_\alpha$. But we have an infinite sum, which is meaningless until we can find a way to interpret it.
Luckily, this isn't really an infinite sum, precisely because of the restriction we made. All but finitely many of the terms are 0. For any $\alpha$ to contribute a nonzero term, it must be a member of both the set of $\alpha_i$'s and the set of $\alpha'j$'s. Let's label these indices by $\beta$, so that ${\beta_k} = {\alpha_i} \cap {\alpha'_j}$. Now we can make sense of the above sum by restricting it to only those indices that contribute nonzero terms, so that $\langle \Psi | \Psi' \rangle = \displaystyle \sum{\beta_k} \langle \Psi_{\beta_k} | \Psi'{\beta_k} \rangle{\beta_k}$. You could just as well sum over either the $\alpha_i$ or $\alpha'_j$ indices, which would still be a finite sum of course, but I prefer making it manifestly (conjugate) symmetric this way.
You have to check at this point that the axioms for an inner product are satisfied by this choice of $\langle \cdot | \cdot \rangle $. It's clearly symmetric up to a complex conjugate, which is what we want for a Hermitian inner product. The fact that it is linear requires some playing around with symbols if you want to prove, though the direction of the proof should be immediately obvious if you've followed the above. Positive-definiteness is inherited from the $\mathcal H_\alpha$. 66