There is no difference between the direct sum and the direct product for finitely many terms, regardless of whether the terms themselves are infinite-dimensional or not. However, they are different in the case of infinitely many terms (and drastically so).
A direct product $\prod_{i = 1}^\infty V_i$ can be thought of as the set of sequences $(v_1, v_2, \ldots)$ where each $v_i \in V_i$, with usual scalar multiplication $\lambda (v_1, v_2, \ldots) = (\lambda v_1, \lambda v_2, \ldots)$ and pointwise addition $(v_1, v_2, \ldots) + (w_1, w_2, \ldots) = (v_1 + w_1, v_2 + w_2, \ldots)$. The direct sum $\bigoplus_{n=1}^\infty V_i$ on the other hand is the same set, with the extra condition that only finitely many terms are nonzero.
The direct sum behaves nicely in terms of bases. If each $V_i$ has some basis set $B_i \subseteq V_i$, then the direct sum $\bigoplus_{n=1}^\infty V_i$ has a basis identified with $\bigsqcup_{i=1}^\infty B_i$. For example, if all $V_i = \mathbb{R}$, then the basis for the direct product is just putting a 1 in the $i$th place for all $i$: $(1, 0, 0, \ldots), (0, 1, 0, 0, \ldots), \ldots$. In particular, if all the $V_i$ are countable dimension, the direct sum is also countable dimension.
With the direct product, this is not the case. It is not so hard to see (I'm sure there are many answers on this site) that the space of sequences of real numbers $\prod_{i=1}^\infty \mathbb{R}$ has uncountable dimension over $\mathbb{R}$.
Finally, there is not that much subtlety in what it means to be a basis of an infinite dimensional space. A basis is a linearly independent subset $B \subseteq V$ such that any vector $v \in V$ may be written as a finite linear combination of basis vectors. This is perhaps the best way to think about the difference between direct sum and product: start trying to write down a system of elements which can express any real sequence as a finite linear combination, and you'll soon see that in many cases, a direct sum may have been what you intended all along.
No, it will not be enough. What does “Let $\beta={v_1,...v_k}$ for $S_0$, $\beta'{v_1',..,v_k'}$ for $S_0^{\bot}$” mean? I have no idea.
Besides, you don't need bases here. If $v\in S^\top$, then, for each $w\in S$, $\langle v,w\rangle=0$. In particular (since $S_0\subset S$), for each $w\in S$, $\langle v,w\rangle=0$, and so $v\in S_0^\top$.
Best Answer
An infinite dimensional inner product space with an inner product (and the additional property of completeness) is called a Hilbert space. It turns out that in a Hilbert space, your statement does not necessarily hold true. In particular a subspace $W$ of a Hilbert space $\mathcal H$ will satisfy $\mathcal H = W \oplus W^\perp$ if and only if $W$ is (topologically) closed.
For example, we necessarily have $\mathcal H = W \oplus W^\perp$ when $W$ is a finite dimensional subspace of $\mathcal H$. To be more specific about where the direct generalization of your proof fails, a subspace $W$ that fails to be closed does not have a Schauder basis.
The study of infinite dimensional vector spaces like this one falls under the domain of functional analysis. If you are interested in a relevant reference, you might want to try reading Kreyszig's Introductory Functional Analysis with Applications, which I find to be "beginner friendly" (yet fairly comprehensive) relative to similar texts.