If you have a vector space $X$ with an inner product $\langle \cdot, \cdot \rangle$, this defines a norm $\|\cdot\|$ by $\|x\|=\sqrt{\langle x, x\rangle}$ (it is a good exercise to prove that this is in fact a norm). Similarly, this defines a metric, $d(x,y)=\|x-y\|$ (it is again a good exercise to prove that this is in fact a metric). This is the case for any inner product space, so yes, an inner product always defines a metric. However, not every metric is defined by an inner product!
A sequence of elements $\{x_n\}$ in $X$ is called a Cauchy sequence if $\|x_n-x_m\|\to0$ as $n,m\to\infty$. An inner product space $X$ is called a Hilbert space if it is a complete metric space, i.e. if $\{x_n\}$ is a Cauchy sequence in $X$, then there exists $x\in X$ with $\|x-x_n\|\to0$ as $n\to\infty$.
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.
Best Answer
Note that, by Parseval's inequality, every element of $H$ is written uniquely as $x=\sum_n\langle x,e_n\rangle e_n$. Conversely, given any sequence $(\lambda_n)\in\ell^2$, the vector $x=\sum_n\lambda_ne_n$ is a well-defined element of $H$ and satisfies $\langle x,e_n\rangle=\lambda_n$.
What you are looking for follows directly by the above and the observation that the sequence $(\frac{1}{n})$ is a sequence in $\ell^2$, i.e. $\sum_n\frac{1}{n^2}<\infty$.
A Hilbert space is an inner product space that is complete under the induced norm $\|\cdot\|:=\sqrt{\langle\cdot,\cdot\rangle}$.