Existence of Schauder bases in separable Hilbert spaces can be proven without any appeal to the axiom of choice, countable or not. Here is the outline.
Let $v_1,v_2,\dots$ be a dense countable subset of the Hilbert space $H$ in question. Let is look at every element $v_i$ of the sequence and look if it is in the linear span of $v_j,j<i$. If it is, we discard it, and if not, we keep it. This way, we get a (possibly finite; for simplicity I'll assume there are infinitely many of them, the proof is the same otherwise) sequence $w_1,w_2,\dots$ of elements of $H$ which form a linearly independent set, and which moreover linearly span a subspace of $H$ containing each $v_i$, so the linear span is in $H$. By applying the Gram–Schmidt process (which is completely constructive, no choice needed) we can conclude there is a sequence $u_1,u_2,\dots$ which is orthonormal and topologically spans $H$. We claim this sequence is a Schauder basis.
Let $v\in H$ be arbitrary. Uniqueness of representation in terms of basis is clear: if $v=\sum_{i=1}^\infty a_iu_i$, then it's easy to see $\langle v,u_i\rangle=a_i$. So we are left with existence. Recall the vectors $v_i$ were dense in $H$, so there is a sequence $v_{i_n}$ convergent to $H$ (no choice needed - e.g. let $i_n$ be the least such that $||v-v_{i_n}||<1/n$. By construction, $v_{i_n}$ is in the linear span of $u_k$, so there is a unique sequence of reals $a_{n,k}$, finitely many of which are nonzero, such that
$$v_{i_n}=\sum_{k=1}^\infty a_{n,k}u_k.$$
The sequence $v_{i_n}$ is Cauchy, hence so is the sequence $a_{n,k}=\langle v_{i_n},a_k\rangle$ for each fixed $k$, thus it converges to some number $b_k$. One can show that then we have $v=\sum_{k=1}^\infty b_ku_k$ (this is very similar to the proof that $\ell_2$ is a Hilbert space), which shows existence. Hence $u_1,u_2,\dots$ is a Schauder basis of $H$.
