Let's denote the tensor product of Hilbert spaces by $\widehat{\otimes}$, the tensor product of vector spaces by $\otimes$.
We have a natural bilinear map $K \times \bigoplus_i H_i \to \bigoplus_i (K \widehat{\otimes} H_i)$ which sends $(a,(h_i))$ to $(a \otimes h_i)$. It extends to a linear map $K \otimes \bigoplus_i H_i \to \bigoplus_i (K \widehat{\otimes} H_i)$. This map is isometric because
$$\langle (a \otimes h_i),(a' \otimes h'_i) \rangle = \sum_i \langle a,a' \rangle \cdot \langle h_i,h'_i \rangle = \langle a,a' \rangle \langle (h_i),(h'_i) \rangle.$$
Hence it extends to an isometric linear map on the completion $K \widehat{\otimes} \bigoplus_i H_i \to \bigoplus_i (K \widehat{\otimes} H_i)$. One checks that its image is dense. The image is also closed since it is complete. Thus it is an isometric isomorphism.
Yes it is true independent of the cardinality of the bases of the vector spaces. Use the universal property: Let $T$ be any vector space over the given field.
A bilinear map $V \times W \to T$ is uniquely determined by the images of the pairs $(v_i,w_j)_{i \in I,j \in J}$, so we get
$$Bil(V \times W, T) = Abb(\{(v_i,w_j), i \in I, j \in J\},T)= Abb(\{v_i \otimes w_j, i \in I, j \in J\},T)$$
By the universal property $V \otimes W$ is the vector-space with the property
$$Hom(V \otimes W,T) = Bil(V \times W, T)$$
, so we obtain
$$Hom(V \otimes W,T) = Abb(\{v_i \otimes w_j, i \in I, j \in J\},T)$$
, which precisely states that $\{v_i \otimes w_j, i \in I, j \in J\}$ is a basis of $V \otimes W$.
Let us precise this:
Given a map in $f \in Abb(\{v_i \otimes w_j, i \in I, j \in J\},T)$, which means that we are given $f(v_i \otimes w_j)$ for all $i \in I, j \in J$, we have to show that this extends uniquely to a map $F:V \otimes W \to T$.
We define the bilinear map $\hat f: V \times W \to T$ by $\hat f(v_i,w_j) := f(v_i \otimes w_j)$. By the universal property, this gives us a unique map $F:V \otimes W \to T, F(v \otimes w)=\hat f(v_i,w_j)=f(v_i \otimes w_j)$. Hence this is the desired unique map.
The uniqueness of $F$ is trivial anyway, because $\{v_i \otimes w_j, i \in I, j \in J\}$ is clearly a set of generators. We only need to show the existence of $F$ here (And this corresponds to the linear independence).
Best Answer
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.