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.
There is an error in your proof, when you write that $\left\{u,\tilde v\right\}\subset B$ “by maximality”. The fact that $B$ is maximal doesn't imply that it contains every orthonormal set. In $\mathbb R^2$, endowed with its usual inner product, $\{(1,0),(0,1)\}$ is a maximal orthonormal set, $\left\{\frac35,\frac45\right\}$ is orthonormal, but $\{(1,0),(0,1)\}\not\supset\left\{\frac35,\frac45\right\}.$
Best Answer
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.