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
An "orthonormal basis" is not the same thing as a basis that is orthonormal, and in particular may not even actually be a basis at all. An orthonormal basis for a Hilbert space $H$ is just a maximal orthonormal subset $B\subseteq H$. Such a subset need not actually span $H$, and so it need not be a basis. It is true that we can write an arbitrary element $x\in H$ as the sum $\sum_{b\in B}\langle x,b\rangle b$, but this is an infinite sum in general and so does not show that $x$ is in the span of $B$.
So, a separable infinite dimensional Hilbert space has a countable orthonormal basis, but no such orthonormal basis is actually a basis so this does not contradict your Theorem 1.