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.
a) If $v$ is orthogonal to $S$, then it is orthogonal to any linear combination of elements of $S$ (this follows from the linearity of the scalar product), i.e. it is orthogonal to the linear span of $S$.
b) If $v$ is orthogonal to some set $S$, then it is orthogonal to its closure (this follows from the continuity of the scalar product, i.e. if $x_n \to x$ and $y_n \to y$, then $\langle x_n, y_n \rangle \to \langle x, y \rangle$).
The meaning of the word "dense" here is exactly the same as in the general topology.
Edit:
3.
This all works in the general case of Banach spaces with Schauder bases (see, e.g., Lindenstrauss-Tzafriri's Classical Banach Spaces, vol 1, Proposition 1.a.3).
Scheme:
Define the projections $P_n(x) = \sum_{k=1}^n \langle x, e_k\rangle e_k$.
Clearly, $P_n u \to u$ for all $u$ in the linear span.
We want to show $P_n x \to x$ for all $x$. For that, it is enough (it's an easy argument using the density condition) to know $\|P_n\| = 1$ for all $n$.
This follows from the linearity of $P_n$, $\|P_n\| \leq n$, and $\|P_n x\| \leq \|P_m x\|$ for all $n < m$ and $x$.
Best Answer
For a Banach space $X$, let $X^*$ denote the dual space (the space of continuous, linear functionals $x^*:X\to \mathbb{K}$, where $\mathbb{K}$ is the scalar field, which is either $\mathbb{R}$ or $\mathbb{C}$). Somewhat suggestively, we denote the action of a functional $x^*$ on the vector $x$ as $\langle x,x^*\rangle$, but we note that $\langle \cdot,\cdot\rangle$ is not an inner product, since the first argument comes from $X$ and the second comes from $X^*$.
In general Banach spaces, the appropriate notion is the notion of biorthogonality. Let $X$ be a Banach space, $A$ a set, $\alpha:A\to X$, and $\beta:A\to X^*$ functions. We say the collection $(\alpha(a), \beta(a))_{a\in A}\in \prod_{a\in A}X\times X^*$ is biorthogonal if for $a,b\in A$, $$\langle \alpha(a),\beta(b)\rangle = \left\{\begin{array}{ll} 1 & : a=b \\ 0 & : a\neq b.\end{array}\right.$$
Suppose we have a Hilbert space with inner product $(\cdot,\cdot):H\times H\to \mathbb{K}$. By (one version of) the Riesz Representation Theorem, there is a canonical conjugate-linear isometric surjection $\phi:H\to H^*$. Then for any set $A$ and any function $\alpha:A\to H$, we can naturally build $\beta:A\to H^*$ as $\beta(a)=\phi(\alpha(a))$. The collection $(\alpha(a))_{a\in A}$ is orthonormal if and only if $(\alpha(a),\beta(a))_{a\in A}=(\alpha(a),\phi(\alpha(a)))_{a\in A}$ is biorthogonal. It is for this reason that I view biorthogonality as the appropriate generalization of the notion of orthonormality. However, it is no longer the case that biorthogonality has some relation to the norm, whereas orthonormality implies the vectors $\alpha(a)$ each have norm $1$.
Let $X$ be a Banach space and let $X^*$ be the dual space (the space of continuous, linear functionals on $X$). A Markushevich basis (or $M$-basis) is a set $A$ together with a pair of functions $\alpha:A\to X$, $\beta:A\to X^*$ such that
Conditions 1,2 yield a biorthogonal system whose first components have dense span in $X$. Condition 3 says that, although the second components need not have dense span in $X^*$, there are still sufficiently many functionals in $\{\beta(a):a\in A\}$ to separate points. We typically use the notation $x_a$ instead of $\alpha(a)$ and $x^*_a$ instead of $\beta(a)$, so the Markushevish basis is denoted $(x_a,x^*_a)_{a\in A}$. We have the following.
There is also the notion of a Schauder basis for $X$. A Schauder basis is a sequence $(x_n)_{n=1}^\infty$ such that for each $x\in X$, there exists a unique scalar sequence $(a_n)_{n=1}^\infty$ such that $x=\sum_{n=1}^\infty a_nx_n:=\lim_N\sum_{n=1}^N a_nx_n$. This implies a property called the approximation property (and even stronger properties). Per Enflo gave the first example of a separable Banach space which lacked the approximation property, and therefore does not admit any Schauder basis. So the $M$-basis result above seems to be the full extent to which a general analogue holds.
As a side note, if we're only concerned with the original norm, and not necessarily a norm which comes from an inner product, we can note that for any Banach space $(X, \|\cdot\|)$, there is an inner product $(\cdot,\cdot)$ on $X$ such that, with $|x|=\sqrt{(x,x)}$ the norm induced by the inner product, both $(X, \|\cdot\|)$ and $(X, |\cdot|)$ are complete. However, there need not be any real connection between $\|\cdot\|$ and $|\cdot|$, so the fact that this dot product exists does nothing to address the original question about $(X, \|\cdot\|)$. That's because neither orthogonality nor normality with respect to $(\cdot,\cdot)$ and $|\cdot|$ have any connection to $\|\cdot\|$. It does not induce any interpretation of "orthogonality" with respect to the original norm $\|\cdot\|$, and $\|\cdot\|$-normality need not be connected to $|\cdot|$ normality. In fact, in the infinite dimensional case, you can guarantee that $\{\|x\|/|x|:0\neq x\in X\}=(0,\infty)$.