Let $(X,\langle\cdot,\cdot\rangle)$ be a real inner product space. Then, there exists a linear isometry from $X$ to its dual $X'$, whose image is dense.
I know that in the case $X$ is complete, the linear isometry we can use is $\Phi_X:X\to X'$ defined by $\Phi_X(x)=\langle\cdot,x\rangle$, and use the Fréchet-Riesz representation theorem.
Now, for general $X$, I want to define the same mapping. It is still a linear isometry, but not surjective anymore since surjectivity requires $X$ to be complete. Therefore, I want to show denseness of $\Phi_X(X)$ in $X'$. However, this is where I get stuck. I want to show that $\overline{\Phi_X(X)}=X'$. The first "$\subseteq$" follows easily, but how do i show "$\supseteq$"?
The answer is in your title: work on the completion of $X$; as $X$ is dense in its completion, its image $\Phi_X(X)$ is dense in $X'$.