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
I will try to answer in a "how to keep things in mind way".
First of all, the most common and most flexible definition of separability is the first: A topological space $X$ is called separable when there exists a countable subset $\{x_n\}_{n=1}^\infty$ that is dense in the space. Now we don't care about topology in general, we care about Hilbert spaces, but the definition is the same.
Using Zorn's lemma, one proves that every Hilbert space admits an orthonormal basis. An orthonormal basis of a Hilbert space $H$ is a set of vectors $(e_i)_{i\in I}$ such that $\|e_i\|=1$ for all $i\in I$, $\langle e_i,e_j\rangle=0$ when $i,j\in I$ and $i\neq j$ and it is true that for each $x\in H$ it is $$x=\sum_{i\in I}\langle x,e_i\rangle e_i$$ in the sense that for each $\varepsilon>0$ we may find a finite set $F_0\subset I$ such that for any finite set $F\subset I$ with $F_0\subset F$ it is $$\bigg{\|}x-\sum_{i\in F}\langle x,e_i\rangle e_i\bigg{\|}<\varepsilon.$$
Don't let this scare you! Once you learn about nets this will be very easy to digest. Just keep in mind that in the case that our index-set $I$ is countable, the sum above is simply a series (in the way that even your cat understands what a series is).
An equivalent definition is this: $(e_i)_{i\in I}$ is an orthonormal basis for $H$ if and only if $\|e_i\|=1$ for all $i$, $\langle e_i,e_j\rangle=0$ for all $i\neq j$ and the following implication is true for any $x\in H$: $$\bigg(\text{ for all i}\in I: \langle x,e_i\rangle=0\bigg)\implies x=0 $$
Anyway. It can be proved that if $(e_i)_{i\in I}$ and $(f_j)_{j\in J}$ are both orthonormal bases for a Hilbert space $H$, then $I$ and $J$ have the same cardinality. We say that this common cardinality is the Hilbert dimension of the Hilbert space.
One can easily convince themselves that a Hilbert space has finite Hilbert dimension if and only if it is finite dimensional in the usual sense.
Now one can also prove the following: a Hilbert space is separable if and only if its Hilbert dimension is (finite or) countably infinite.
Hint on how to prove the last thing: Obviously finite dimensional spaces are separable. If a space is separable, start from a dense sequence, pass to a dense linearly independent subsequence and apply Gram-Schmidt. Conversely, if a space has countably infinite Hilbert dimension, then the $\mathbb{Q}$-span of a countable orthonormal basis is a dense, countable set.
A good reference for all this is Conway's first Chapter. The exercises are difficult though, be warned about that!