Assume $X$ to be a separable infinite dimensional Hilbert-space and $Y\subset X$ to be a not necessarily closed subspace. Then $Y$ is also separable as subspaces of separable metric spaces are always separable. This again means that $Y$ has a countable basis $\left\lbrace y_k:\ k \in \mathbb{N} \right\rbrace$. Now using Gram-Schmidt (for the infinite dimensional case) we can transform $\left\lbrace y_k:\ k \in \mathbb{N} \right\rbrace$ into an orthonormal basis $\left\lbrace z_k:\ k \in \mathbb{N} \right\rbrace$, where orthonormal basis means a set of vectors $z_k$, orthonormal wrt. the inner product on $X$ and such that $\overline{\text{lin}}\left\lbrace z_k:\ k \in \mathbb{N} \right\rbrace=Y$ and $\overline{\text{lin}}$ denotes the closure of the linear hull.
Is the above reasoning correct, i.e. does any subspace of a separable Hilbert-space have a countable orthonormal basis in this sense? What confuses me is that in general I can find this statement only for $Y$ being a closed subspace, but I think by these argument it should also hold true for general $Y$.
Yes, the problem is precisely that $Y$ is not a Hilbert space if it's not closed.
Note that $\overline{\text{lin}}S$ is closed by definition, so if $Y$ is not closed we cannot have $Y=\overline{\text{lin}}S$, hence $Y$ cannot have a basis by your definition.