Visualize the proof of $\mathcal{H} = M \oplus M^{\perp}$, where $\mathcal{H}$ Hilbert space, $M$ closed subspace


Put $m=(a,b,c)$. The set $K$ is a copy of a set $-M$ shifted by $x$, that is $K=\{x-\mu m:\mu\in\Bbb R\}$ is a line. By the definition, $x_2$ is a point of $K$ closest to the origin $O=(0,0,0)$. Looking at the plane spanned by $O$ and $K$, we see that $x_2$ is the base of the altitude dropped from $O$ to $K$, that is $x_2\in K$ and $\langle x_2, m\rangle=0$. Thus $x_2=x-\mu m=(u,v,w)-\mu (a,b,c)$, with $\mu=\tfrac{ua+vb+wc}{a^2+b^2+c^2}=\frac{\langle m,x\rangle}{\|m\|^2} .$

And what is the geometric intuition of choosing $\lambda$ that way?

In order to obtain the best from the inequality $\| x_2 \|^2 \le \| x_2 - \lambda y \|^2$ we pick $\lambda$ which minimizes its right hand side. For this we need $x_2 - \lambda y$ be the base of the altitude dropped from $O$ to the line $\{x_2-\lambda y:\lambda\in\Bbb R\}$. Similarly to the above we see that this is attained when $\lambda := \frac{\langle x_2, y \rangle}{\langle y, y, \rangle}$.