Existence: Let $d = \inf_{y \in M} \|x - y\|$. Choose a sequence $(m_n)$ in $M$ such that $\|x - m_n\| \to d$ as $n \to \infty$.

We show $(m_n)$ is Cauchy using the parallelogram law: $\|m_n - m_k\|^2 = \|(x - m_k) - (x - m_n)\|^2$ $= 2\|x - m_k\|^2 + 2\|x - m_n\|^2 - \|2x - m_k - m_n\|^2$ $= 2\|x - m_k\|^2 + 2\|x - m_n\|^2 - 4\left\|x - \frac{m_k + m_n}{2}\right\|^2$

Since $M$ is a subspace, $(m_k + m_n)/2 \in M$, so $\left\|x - \frac{m_k + m_n}{2}\right\|^2 \geq d^2$

As $n, k \to \infty$, we have $\|x - m_n\|^2, \|x - m_k\|^2 \to d^2$, therefore $\|m_n - m_k\|^2 \leq 2d^2 + 2d^2 - 4d^2 = 0$

So $(m_n)$ is Cauchy. Since $H$ is complete and $M$ is closed, $m_n \to m \in M$. By continuity of the norm, $\|x - m\| = d$.

Orthogonality Characterization: For any $y \in M$ and $\lambda \in \mathbb{K}$, we have $m + \lambda y \in M$, so $\|x - m\|^2 \leq \|x - m - \lambda y\|^2 = \|x - m\|^2 - 2\text{Re}(\lambda \langle x - m, y \rangle) + |\lambda|^2 \|y\|^2$

This gives $2\text{Re}(\lambda \langle x - m, y \rangle) \leq |\lambda|^2 \|y\|^2$ for all $\lambda$.

Choosing $\lambda = t \langle x - m, y \rangle$ for small $t > 0$ yields $2t |\langle x - m, y \rangle|^2 \leq t^2 |\langle x - m, y \rangle|^2 \|y\|^2$. Dividing by $t$ and letting $t \to 0$ gives $\langle x - m, y \rangle = 0$ for all $y \in M$.

Uniqueness: If $m_1, m_2 \in M$ both satisfy $\|x - m_1\| = \|x - m_2\| = d$, then by the parallelogram law: $\|m_1 - m_2\|^2 = 2\|x - m_1\|^2 + 2\|x - m_2\|^2 - 4\left\|x - \frac{m_1 + m_2}{2}\right\|^2$ $\leq 2d^2 + 2d^2 - 4d^2 = 0$

Therefore $m_1 = m_2$.