Given a non-empty set $D$ and $\mathbb{K}\in\lbrace\mathbb{R}, \mathbb{C}\rbrace$, let $H$ be a subspace of $\mathbb{K}^D$ with an inner product $\langle\cdot, \cdot\rangle$. It is given that a map $K: D\times D\rightarrow\mathbb{K}$ exists with the property that $K(\cdot, t)\in H$ for all $t\in D$ and $\langle f, K(\cdot, t)\rangle=f(t)$ for alle $f\in H, t\in D$ (i.e., $K$ is a reproducing kernel of $H$). Note that $H$ may or may not be complete (i.e., may or may not be a Hilbert space).
Is it true that $\text{span}\lbrace K(\cdot, t): t\in D\rbrace$ is dense in $H$?
It is pretty easy to show that this is true if $H$ is a Hilbert space. Namely, we have $\text{span}\lbrace K(\cdot, t): t\in D\rbrace^\perp=\lbrace 0\rbrace$ immediately from the reproducing property of $K$ and if $H$ is a Hilbert space, this means $$\overline{\text{span}\lbrace K(\cdot, t): t\in D\rbrace}=\text{span}\lbrace K(\cdot, t): t\in D\rbrace^{\perp\perp}=\lbrace 0\rbrace^\perp=H.$$
However, this argument breaks down if $H$ is not a Hilbert space. In an attempt to prove that the statement is also true for $H$ not a Hilbert space, I have tried to consider a completion $\tilde{H}$ of $H$ and then argue in $\tilde{H}$, but to no success.
I am getting the feeling, however, that $\text{span}\lbrace K(\cdot, t): t\in D\rbrace$ may not even be dense in $H$ if $H$ is not complete, but could not come up with a suitable example either.
Can anyone give a proof that the statement still holds in non-complete inner product spaces $H$ or an example of such a space where the statement is false?