Every partially defined isometry can be extended to a isometry

Solution 1:

Infinite-dimensional case

False. Consider the backward shift operator on $\ell^2$: $$S^*(x) = (x_2,x_3,\dots)$$ which is an isometry of the subspace $\{x:x_1=0\}$ onto $\ell^2$. It can't be extended further since it's already surjective.

Complex case

False. For example, there is no unitary transformation $T:\mathbb{C}\to \mathbb{C}$ such that $T(1)=1$ and $T(i)=-i$. The issue is that as long as Euclidean metric is concerned, $\mathbb{C}^n$ is just $\mathbb{R}^{2n}$. But the unitary group $U(n)$ is much smaller than $O(2n)$, due to the $\mathbb{C}$-linearity requirement.

Real case

True. A reference is Theorem 11.4 in Embeddings and Extensions in Analysis by J.H. Wells and L.R. Williams. Terminological note: they say that a pair of spaces $(X,Y)$ has the isometric extension property if every isometry from a subset of $X$ into $Y$ can be extended to an isometry of $X$ into $Y$.