= Solution
Let $d=\inf_{v\in V}\|f-v\|$ and choose $v_n\in V$ with $\|f-v_n\|\to d$. The parallelogram identity gives
$$
\|v_n-v_m\|^2
=2\|f-v_n\|^2+2\|f-v_m\|^2-4\left\|f-\frac{v_n+v_m}{2}\right\|^2
\leq2\|f-v_n\|^2+2\|f-v_m\|^2-4d^2\longrightarrow0.
$$
Completeness and closedness give $v_n\to v\in V$, attaining the distance. Put $w=f-v$. For $u\in V$ and real $t$, minimality gives $\|w-tu\|^2\geq\|w\|^2$, so $\operatorname{Re}\langle w,u\rangle=0$. In a complex <Hilbert space>, also replace $u$ with $iu$ to get the imaginary part zero. Therefore
$$
\boxed{f=v+w,\qquad v\in V,\quad w\perp V.}
$$
The decomposition is unique since $V\cap V^\perp=\{0\}$. This proves the required <orthogonal projection> result rather than assuming it before the averaging argument.
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