I think you are getting confused with the notation here. For each of your $x_n$, its coordinates are vectors in $H_n$; so "$1$" makes no sense there.
Note that we cannot expect the construction to always give a non-separable Hilbert space. Because we can take $H_n=\mathbb C$ for all $n$, and then $\prod_nH_n=\ell^2(\mathbb N)$, which is separable.
Note also that you never defined what the inner product in the direct sum is, but your condition on the norms suggests that it is the canonical one, $ \langle x,y\rangle = \sum_n\langle x_n,y_n\rangle. $
To make up a basis of $\prod_nH_n$, the natural way is to use bases from each of the $H_n$. So, for each $n$, fix an orthonormal basis $B_n=\{e_{k,n}\}_{k\in K_n}$ of $H_n$.
Let $ B=\{\,x\in\prod_nH_n:\ \exists m\text{ with }x_m\in B_m\text{ and }x_r=0\text{ if }r\ne m\} $ It is clear that $B$ is an orthonormal set. Now suppose that $x\in B^\perp$. Then, for any $m$ and any $k\in K_m$, $ 0=\langle x,e_{k,m}\rangle=\langle x_m,e_{k,m}\rangle. $ As $k\in K_m$ was arbitrary, $x_m=0$; as $m$ was arbitrary, $x=0$. So $B$ is total, and it is thus a basis.
Now, if all the $H_n$ are separable, then $K_n$ is countable for all $n$. As the family $H_1,H_2,\ldots$ is countable, we have $ B=\bigcup_{m\in\mathbb N}\{x:\ x_m\in B_m\text{ and zero elsewhere }\}. $ As each $B_m$ is countable, $B$ is a countable union of countable sets, and is itself counatble. So $H$ is separable.
If any of the spaces $H_1,H_2,\ldots$ is non-separable, or if the family $\{H_n\}$ is uncountable, then $\prod H_n$ will be non-separable.