It's been a while since I was made to look at the proof rather than just quote it, but IIRC the gist of the RRT for Hilbert spaces, is the bijection between closed hyperplanes (=closed codimension $1$ subspaces) in a Hilbert space $H$ and the lines orthogonal to each, and the fact that this can be set up so as to be conjugate-linear. This in turn is based -- I think -- on the fact that for each $x in H$ and each closed subspace $V$ there is a unique point in $V$ closest to $x$.
If you wanted to look at the (continuous) dual of an inner product space $E$, then the above reasoning suggests to me that completion of $E$ is going to enter the picture somehow. For if $psi$ is a continuous linear functional on $E$, we want to consider $ker psi$ and then associate to it a choice of normal vector, but I'm not sure we can show that a suitable choice exists without using completeness.
(There exist plenty of codimension 1 dense subspaces in incomplete i.p. spaces, of course: equip $C[0,1]$ with the inner product given by integrating along $[0,1]$, i.e. the $L^2[0,1]$ inner product, and consider the subspace of $C[0,1]$ consisting of all those functions in it which vanish at $0$. So in the setting above, the continuity of ψ has to get used in the proof that the dual of $E$ is a Hilbert space.)
I take your point that perhaps there is a way to show that the dual of $E$ is a Hilbert space, which doesn't start by completing $E$. But one may end up constructing some kind of abstract completion anyway.
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