= Solution
Put $e=x-u^*$. If $u^*$ minimizes distance, then for every $v\in\mathcal U_n$ and every real $t$,
$$
\|e-tv\|^2=\|e\|^2-2t\operatorname{Re}(e,v)+t^2\|v\|^2\geq\|e\|^2.
$$
Both signs of arbitrarily small $t$ force $\operatorname{Re}(e,v)=0$. In a complex <inner product space>, apply the same argument to $iv$ to obtain the imaginary part too. Conversely, if $e$ is orthogonal to the subspace, the <Pythagorean theorem for inner product spaces> gives
$$
\|x-u\|^2=\|x-u^*\|^2+\|u-u^*\|^2\qquad(u\in\mathcal U_n).
$$
Thus \b[<orthogonality> characterizes the unique best approximation]:
$$
\boxed{u^*\text{ is best}\iff(x-u^*,v)=0\quad(v\in\mathcal U_n).}
$$
Finite dimension also guarantees existence by solving the invertible <Gram matrix> normal equations; completeness of the ambient <inner product space> is unnecessary.
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