This is actually an if and only if statement. For future students looking for a full answer to this question, I am posting a full proof below:
Let $H$ be an infinite-dimensional Hilbert space. Show that $H$ has a countable orthonormal basis if and only if $H$ has a countable dense subset.
First let us assume that $H$ has a countable orthonormal basis $\{e_i\}$. Then any $x$ can be uniquely written as
$$x=\sum\limits_{i=1} c_ie_i \quad \text{where} \quad c_i= \langle x,e_i \rangle$$
Recall that $S:=\mathbb{Q}+\mathbb{Q}i$ is a countable dense subset of $\mathbb{C}$. Now for every $n \in \mathbb{N}$, consider the following subset of $H$:
$$A_n=\left\{\sum\limits_{i=1}^n s_ie_i \quad \text{where} \quad s_i \in S \, \, \forall i\right\}.$$
Being a finite union of countable sets, each $A_n$ is countable. Then define
$$A:= \bigcup_{n=1}^\infty A_n$$
Being a countable union of countable sets, $A$ is countable. Let us show that $A$ is a dense subset of $H$; its being countable will imply separability of $H$.
Let $x\in H$. Then
$$x=\sum\limits_{i=1}^\infty c_ie_i \quad \text{where} \quad c_i=\langle x,e_i\rangle \in \mathbb{C}.$$
Since this infinite sum is convergent in the norm of $H$, fixing an arbitrary $\epsilon >0$, we can find a big enough $N$ for which
$$\left\| \sum\limits_{i=N+1}^\infty c_ie_i\right\|< \frac{\epsilon}{2}.$$
Also, Since $S$ is a dense subset of $\mathbb{C}$, for every $i \leq N$, we can find $s_i\in S$ such that
$$|c_i-s_i|<\frac{\epsilon}{2^{i+1}}.$$
Now consider the following element
$$x_N=\sum\limits_{i=1}^N s_ie_i \in A.$$
We know that $\sum\limits_{i=N+1}^\infty c_ie_i$ is the difference of two elements of $H$ and thus is in $H$ and therefore by using triangle inequality and Perseval's inequality, we have
\begin{equation}
\begin{split}
\|x-x_N\|
& =\left\|\sum\limits_{i=1} c_ie_i-\sum\limits_{i=1}^N s_ie_i\right\|\\
& =\left\|\sum\limits_{i=1}^N (c_i-s_i)e_i+\sum\limits_{i=N+1} c_ie_i\right\|\\
& \leq \left\|\sum\limits_{i=1}^N (c_i-s_i)e_i\right\|+\left\| \sum\limits_{i=N+1} c_ie_i\right\|\\
& \leq \sum\limits_{i=1}^N \left|(c_i-s_i)\right|+\frac{\epsilon}{2}\\
& < \sum\limits_{i=1}^N \frac{\epsilon}{2^{i+1}}+\frac{\epsilon}{2}\\
& \leq \sum\limits_{i=1}\frac{\epsilon}{2^{i+1}}+\frac{\epsilon}{2}\\
& = \epsilon
.
\end{split}
\end{equation}
Therefore, $x_n$ is an element of the set $A$ that is in the ball of radius $\epsilon$ around $x$. Since both $x$ and $\epsilon$ were arbitrary, this implies that $A$ is dense in $H$.
Conversely, assume that $H$ has a countable dense subset $\{a_j\}$, where $j \in \mathbb{N}$. Let $\{e_i\}_{i \in I}$ be an orthonormal basis of $H$ (we know that such a basis exists by Zorn's lemma).
Proceeding by contradiction, let us assume that the orthonormal basis is uncountable. This implies that for any $e_n \neq e_m$, $n,m \in I$, by orthogonality we have
\begin{equation}
\begin{split}
\|e_n-e_m\|^2
& =\langle e_n-e_m, e_n-e_m \rangle
\\
& =\langle e_n, e_n \rangle+\langle e_m, e_m \rangle-2\text{Re}\big(\langle e_n, e_m \rangle\big)\\
& =\|e_n\|+\|e_m\|\\
& =2\\
\end{split}
\end{equation}
Therefore, any two elements of our orthonormal basis are $\sqrt{2}$ apart. Now for all $i \in I$, consider the following balls:
$$B\left(e_i, \frac{1}{2}\right)$$
Each of such balls has diameter less than $1$. Thus, each one of them can contain at most one element of the basis, namely only the center itself. Also, these ball are disjoint, since if there exists an element in two balls, then by the triangle inequality the distance between the centers of the balls is less than $1$, which is a contradiction. Since $\{a_j\}$ is a dense subset, it has to have at least one element in each such ball. Since by the above remarks the balls are disjoint, we have a surjective function from some $\{a_j\}$ to the balls. However, our balls are indexed by an uncountable set. Thus, there has to be at least an uncountable amount of $a_j$, a contradiction. Thus, any orthonormal basis has to be countable.
Best Answer
As mentioned in the comments, your use of the term "orthonormal basis" is unconventional : Usually a complete orthonormal set is an orthonormal basis. However, if the space is infinite dimensional, then such a set cannot be a Hamel basis :
Any orthonormal set is linearly independent: You don't need uncountable sums here! If $A$ is an orthonormal set and $\{e_1,e_2,\ldots, e_n\} \subset A$ and $\{\alpha_1,\alpha_2,\ldots, \alpha_n\}$ scalars, then $$ \sum_{i=1}^n \alpha_i e_i = 0 \Rightarrow \alpha_j = \langle \sum_{i=1}^n \alpha_i e_i, e_j\rangle = 0 \quad\forall 1\leq j\leq n $$ Hence, $A$ is linearly independent.
If $H$ is a finite dimensional Hilbert space, then every complete orthonormal set has $\dim(H)$ elements. Since they are linearly independent by (1), they must form a Hamel basis.
If $H$ is infinite dimensional and $A$ is an orthonormal basis, then $A$ must be infinite (as before). So choose a countable subset $\{e_n : n\in \mathbb{N}\} \subset A$. Then the series $$ \sum_{n=1}^{\infty} \left\| \frac{e_n}{n^2} \right\| $$ converges. Since $H$ is complete, there is a vector $x\in H$ such that $$ x = \sum_{n=1}^{\infty} \frac{e_n}{n^2} $$ Check that $x$ cannot be expressed as a finite linear combination of elements of $A$. Hence, $A$ cannot be a Hamel basis.