= Solution
Write $r=x-u^*$. If $u^*$ is a best approximation, then for every $v\in\mathcal U_n$ and real $t$, the competitor $u^*+tv$ satisfies
$$
\|r-tv\|^2-\|r\|^2=-2t\operatorname{Re}(r,v)+t^2\|v\|^2\ge0.
$$
The linear coefficient must be zero, since otherwise choosing sufficiently small $t$ of the appropriate sign makes this expression negative. Hence $\operatorname{Re}(r,v)=0$. In a real <inner product space> this is the desired orthogonality; in a complex <inner product space>, apply the same argument to $iv$ to show that the imaginary part also vanishes. Thus $(r,v)=0$ for all $v\in\mathcal U_n$.
Conversely, suppose this orthogonality holds. Every competitor $u\in\mathcal U_n$ has $u-u^*\in\mathcal U_n$. The <Pythagorean identity> then gives
$$
\|x-u\|^2=\|r-(u-u^*)\|^2=\|r\|^2+\|u-u^*\|^2\ge\|r\|^2.
$$
Equality forces $u=u^*$. We have proved the characterization, as well as uniqueness:
$$
\boxed{u^*\text{ is best}\iff (x-u^*,v)=0\text{ for every }v\in\mathcal U_n.}
$$
This is the <orthogonal projection onto a finite-dimensional subspace>. Completeness of the ambient space is not needed. Existence follows, for example, by solving the invertible <Gram matrix> system in a <basis> of the finite-dimensional <linear subspace>.
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