Here is a very general, abstract construction. Let $X$ be an incomplete seperable inner product space and let $Y$ be its completion. Since $X$ is incomplete, there is some $y \in Y \setminus X$. If we had $y \perp X$, this would imply (by density) that $y \perp Y$ and hence $y = 0\in X$, a contradiction. Hence, $y\not\perp X$. Thus, by renormalizing, there is $x_0 \in X$ with $\langle x_0, y\rangle =1$.
Now, the function $\varphi : Y \to \Bbb{K}, x \mapsto \langle x, y\rangle$ is a bounded functional on $Y$ and thus restricts to a bounded linear functional on $X$. Let
$$
M := \{x \in X \mid \varphi(x) = 0\} = X \cap \rm{ker}\,\varphi
$$
and note that $M$ is closed in $X$.
Furthermore, $M \subset \rm{ker}\,\varphi$ is dense. Indeed, let $z \in \rm{ker}\,\varphi$. Then there is a sequence $(x_n)_n$ in $X$ with $x_n \to z$. Hence, $\varphi(x_n) \to \varphi(z) = 0$. Now let $x_n ' := x_n - \varphi(x_n) x_0$. Then $x_n ' \in M$ and it is easy to see $x_n \to z$.
Now, choose a countable dense set $(m_n)_n$ in $M$ and orthonormalize it using the Gram Schmidt procedure, producing an orthonormal set $(x_n)_n$ in $M$ with
$$
\overline{\rm{span}(x_n)_n} = \overline{\rm{span}(m_n)_n} = M.
$$
Here, we take the closure in $X$. Note that the above is indeed true (we don't get all of $X$), since $M$ is closed in $X$. Thus, $(x_n)_n$ is not an orthonormal basis of $X$.
But $(x_n)_n$ is complete in your sense: Since if $x \in X$ satisfies $x \perp x_n$ for all $n$, then (by density) $x \perp M$. But as saw above, $M$ is dense in $\rm{ker}\,\varphi$, so that $x \perp \rm{ker}\,\varphi$ (where we consider $x$ as an element of the completion $Y$).
But it is easy to see $\rm{ker}\,\varphi = (\rm{span}(y))^\perp$, so that (recall that $Y$ is complete) we get $x \in ((\rm{span}(y))^\perp)^\perp = \rm{span}(y)$. Since $y \in Y \setminus X$ and $x \in X$, this implies $x=0$.
One can now certainly make this construction concrete by choosing e.g. $X = \Bbb{K}[X]$ and $Y = L^2([0,1])$ or $X = \ell_0$ (the finitely supported sequences) and $Y = \ell^2$, but I leave this to you as an exercise.
Think of the trigonometric Fourier series in $C^0([0,2\pi])$ the space of all continuous functions with the $L^2$-norm.
$$
||f||_2:=\left(\int_0^{2\pi} |f(x)|^2\,{\rm d}x\right)^{1/2}
$$
This is not complete inner product space in the Cauchy sense, but it is known that any vector (continuous function) in $C^0([0,2\pi])$ has a Fourier expansion (closed). And the zero vector is the only one with zero coefficeints (complete).
Note the closedness and completeness of the the orthonormal systems have nothing to do with topology. Actaully, the set of all (finite) linear combinations of trigonometric function $\cos nx$, $\sin nx$ and some constant is neither complete (Cauchy sense) nor closed (not every limit point is in it). See this.
Best Answer
Let $\{ e_{\alpha} \}_{\alpha\in\Lambda}$ be an orthonormal subset of a Hilbert space $H$, and let $x\in H$ be given. Then the following are equivalent:
The orthonormal subset $\{ e_{\alpha} \}$ is complete if every $x\in H$ satisfies the above. If you let $M$ denote the subspace of all finite linear combinations of elements of $M$, then the following are equivalent:
$\{ e_{\alpha} \}$ is complete.
$\sum_{\alpha}\langle x,e_{\alpha}\rangle e_{\alpha}=x$ for all $x\in H$.
$\sum_{\alpha}\langle x,e_{\alpha}\rangle|^2 =\|x\|^2$ for all $x\in H$.
The subspace $M$ spanned by the $\{e_{\alpha}\}$ is dense in $X$.
$\langle x,e_\alpha\rangle =0$ for all $\alpha$ iff $x=0$.
You can abstract to looking only at a subspace, which leads you to consider a subspace $M$ in a Banach space $X$ that is generated by taking finite linear combinations of the set of elements $\{ x_{\alpha} \}_{\alpha\in\Lambda}$. The closure $\overline{M}$ of $M$ consists of every element $x\in X$ such that, for every $\epsilon > 0$, there is a finite linear combination $\sum_{n=1}^{k}\mu_n x_{\alpha_n}$ such that $\|\sum_{n=1}^{k}\mu_n x_{\alpha_n} -x\|< \epsilon$. In the case of orthonormal sets $\{ e_{\alpha} \}_{\alpha\in\Lambda}$, any such $x$ must be equal to $\sum_{\alpha\in\Lambda}\langle x,e_{\alpha}\rangle e_{\alpha}$. But in a Banach space, there may be no way to write the element as such a sum. If the set $\{ x_{\alpha} \}_{\alpha\in\Lambda}$ is total in a Banach space, there is a sequence of finite sums of the $x_{\alpha}$ that converges to $x$, but that does not necessarily give $x$ as an infinite sum.
Having a sum representation in a Banach space $X$ is usually posed in terms of a countable Schauder basis $\{ x_n \}$ where, for every $x\in X$, one assumes the existence of unique constants $\alpha_n$ such that $x = \sum_{n} \alpha_n x_n$. Every complete orthonormal basis of a Hilbert space is a Schauder basis for that Hilbert space. $\ell^1$ has a Schauder basis consisting of the sequences $\{ 1,0,0,0,\cdots \}$, $\{ 0,1,0,0,0,\cdots \}$, $\{ 0,0,1,0,0,\cdots \}$, etc..
Not every Banach space has a Schauder basis. For example, $L^1[-\pi,\pi]$ does not have a Schauder basis. Even though you might guess the Fourier basis $\{ e^{inx} \}_{n=-\infty}^{\infty}$ to be a Schauder basis for $L^1$, it isn't because the Fourier series does not converge in $L^1$ to an arbitrary $f\in L^1$. However, $\{ e^{inx} \}_{n=-\infty}^{\infty}$ is total in $L^1$ because the finite span $M$ of these functions is dense in $L^1$. Of course $\{ e^{inx} \}$ is a Schauder basis for $L^2$ because it's an orthonormal basis of the Hilbert space $L^2$.
For a Banach space $X$, a set of vectors $\{ x_\alpha \}_{\alpha\in\Lambda}$ is total if the subspace $M$ spanned by these vectors is dense in $X$. Equivalently, every $x$ can be approximated arbitrarily closely by a finite linear combination of the $x_\alpha$. Another equivalent is that the only $x^* \in X^*$ for which $x^*(x_\alpha)=0$ for all $\alpha$ is $x^* = 0$. This condition extends (5) for the Hilbert space case because every $x^*$ on a Hilbert space has the representation $x^*(y)=\langle y,x\rangle$ for a unique $x$.