 Solution

Set $u={\sum }_{k=1}^n⟨x,e_k⟩e_k$, with the convention that the inner product is linear in its first argument. Orthonormality gives $⟨x-u,e_j⟩=0$, hence $x-u\perp U_n$. For any $v\in U_n$ the Pythagorean identity yieldsThus this $u$ is the best approximation, and uniqueness gives $\boxed{P_{n}x=\sum _{k=1}^n⟨x,e_k⟩e_k}$. No completeness assumption on $H$ is necessary for this finite-dimensional construction.