 Solution

For $x\in X$, choose $x_m\in D$ with $x_m\to x$ and defineThe sequence on the right is Cauchy becauseand it converges because $Y$ is complete. The same estimate shows that the limit is independent of the approximating sequence, is linear, and satisfiesIts norm is at most $\Vert T\Vert$, while restriction to $D$ gives the reverse inequality, soAny two continuous extensions agree on the dense set $D$ and hence everywhere, proving uniqueness. This is the extension of a bounded linear operator from a dense subspace.

Completeness of $X$ was never used, so the result remains true when $X$ is incomplete. Completeness of $Y$ is essential. For example, with $D=c_{00}\subset X={\ell }^2$, let the codomain be the incomplete normed space $Y=c_{00}$ with the ${\ell }^2$ norm and let $T$ be the identity. An extension would have to send every limit in ${\ell }^2$ to itself and therefore could not take values in $c_{00}$. This is the failure of dense-subspace extension into an incomplete codomain.