[proofplan] Use an orthonormal basis of the finite-dimensional subspace. The displayed formula gives the candidate projection. Orthogonality gives uniqueness, and the Pythagorean identity gives the minimizing property. [/proofplan] [step:Construct the candidate] Since $M$ is finite-dimensional, choose an orthonormal basis $e_1,\ldots,e_m$ of $M$. Define \begin{align*} p=\sum_{i=1}^m (x,e_i)_V e_i. \end{align*} Then $p\in M$. For each $j$, \begin{align*} (x-p,e_j)_V &=(x,e_j)_V-\sum_{i=1}^m (x,e_i)_V(e_i,e_j)_V \\ &=(x,e_j)_V-(x,e_j)_V=0. \end{align*} By linearity, $(x-p,m)_V=0$ for every $m\in M$, so $x-p\in M^\perp$. [/step] [step:Prove uniqueness] Suppose $q\in M$ and $x-q\in M^\perp$. Then $p-q\in M$. Also \begin{align*} p-q=(x-q)-(x-p), \end{align*} where both $x-q$ and $x-p$ are orthogonal to $M$. Hence $p-q\in M^\perp$. Thus $p-q\in M\cap M^\perp$, so \begin{align*} \|p-q\|_V^2=(p-q,p-q)_V=0. \end{align*} Therefore $p=q$. [/step] [step:Prove minimality] For any $m\in M$, write \begin{align*} x-m=(x-p)+(p-m). \end{align*} The first term is in $M^\perp$ and the second is in $M$, so the two terms are orthogonal. Hence \begin{align*} \|x-m\|_V^2=\|x-p\|_V^2+\|p-m\|_V^2\ge \|x-p\|_V^2. \end{align*} Taking square roots and then the infimum over $m\in M$ gives \begin{align*} \inf_{m\in M}\|x-m\|_V\ge \|x-p\|_V. \end{align*} The reverse inequality follows by taking $m=p$, so equality holds. [/step]