Is there another proof of $V=U+U^\perp$ especially when $V$ is infinite-dimensional? I'm studying Linear Algebra Done Right, the author gave a proof of the theorem(see the figure). However, I personally not quite like the proof, because in a step he use the fact that the projection of $v$ on $U$ is $\langle v,e_1\rangle e_1+\cdots+\langle v,e_m\rangle e_m$. However, this is known by the hindsight. Until the theorem appears, the author didn't(and was unable to) give the proof of this fact. I know that this is rigorous, but I feel that I like the proof without using it. If the vector space $V$ is finite-dimensional, I know that there are many proofs of this theorem, and I can give my own one. However, those proofs for finite-dimensional I know, can not be directly extend to the infinite-dimensional case. Are there another proofs for the case $V$ is infinite-dimensional?

PS:  As you might want to know, this is the definition of projection in that book, after the theorem.

 A: Here's the proof given in Conway's A Course in Functional Analysis, 2nd ed. (Theorems 2.5 and 2.6).

Step 1: If $\mathscr{H}$ is a Hilbert space, $K$ a closed, convex subset of $\mathscr{H}$, and $h \in \mathscr{H}$ then there is a unique point $k_0 \in K$ such that $$ \lVert h - k_0 \rVert = \operatorname{dist}(h,K) \equiv \inf \{ \lVert h - k \rVert : k \in K \}. $$

Proof. We start by noting that by translating $K$ to $K - h \equiv \{k - h : k \in K\}$ we may assume $h = 0$. Next, let $d = \operatorname{dist}(0,K)$. Pick some sequence of points $(k_n) \in K$ such that $\lVert k_n \rVert \to d$. We will use convexity to show that $(k_n)$ is a Cauchy sequence and then use the closedness of $K$ to conclude that $k_n \to$ some $k_0 \in K$.
We note that, by the Parallelogram Law, $$\left\lVert \frac{k_n - k_m}{2} \right\rVert^2 = \frac{\lVert k_n \rVert^2 + \lVert k_m \rVert^2}{2} - \left\lVert \frac{k_n + k_m}{2} \right\rVert^2. $$
Now, since $K$ is convex, $\frac12 (k_n + k_m) \in K$ (you can also take this as the definition of convexity if you are unfamiliar). Therefore, since $d = \operatorname{dist}(0,K)$, we have $\lVert \frac12 (k_n + k_m) \rVert^2 \ge d^2$.
Now let $N$ be such that $n \ge N$ implies $\lVert k_n \rVert^2 < d^2 + \frac14 \varepsilon^2$. We can do this since $\lVert k_n \rVert \to d$. It follows that
$$ \left\lVert \frac{k_n - k_m}{2} \right\rVert^2 < \frac{2d^2 + \frac{1}{4}\varepsilon^2}{2} - d^2 = \frac14\varepsilon^2. $$
Therefore, $\lVert k_n - k_m \rVert < \varepsilon$. I.e. the sequence $(k_n)$ is Cauchy.
Since $\mathscr{H}$ is complete, and $K$ is closed, there is a limit point $k_0 \in K$ such that $k_n \to k_0$. Since the norm is continuous, $$\lVert k_0 \rVert = \lim_{n \to \infty} \lVert k_n \rVert = d. $$
This shows existence.
For uniqueness, suppose $k_0'$ is a second point in $K$ such that $\lVert k_0' \rVert = d$. Then
$$ d \le \lVert \tfrac12 (k_0 + k_0') \rVert \le \tfrac12 (\lVert k_0 \rVert + \lVert k_0' \rVert) = d. $$
Hence, $\lVert \frac12 (k_0 + k_0') \rVert = d$.
Applying the Parallelogram Law,
$$ \left\lVert \frac{k_0 - k_0'}{2} \right\rVert^2 = \frac{\lVert k_0 \rVert^2 + \lVert k_0' \rVert}{2} - \left\lVert \frac{k_0 + k_0'}{2} \right\rVert^2 = \frac{d^2 + d^2}{2} - d^2 = 0. $$
Therefore $k_0' = k_0$.

Step 2: If $\mathscr{M}$ is a closed linear subspace of $\mathscr{H}$, $h \in \mathscr{H}$ and $f_0 \in \mathscr{M}$ is the unique element such that $\lVert h - f_0 \rVert = \operatorname{dist}(h, \mathscr{M}$ then $h - f_0 \perp \mathscr{M}$. Conversely, if $f_0 \in \mathscr{M}$ is such that $h - f_0 \perp \mathscr{M}$ then $\lVert h - f_0 \rVert = \operatorname{dist}(h, \mathscr{M})$.

Proof. Take $f \in \mathscr{M}$. Then $f_0 + f \in \mathscr{M}$ so we have
$$ d^2 = \lVert h - f_0 \rVert^2 \le \lVert h - (f_0 + f) \rVert^2 = \lVert (h - f_0) - f \rVert^2 = \lVert h - f_0 \rVert^2 - 2\operatorname{Re}\langle h - f_0, f \rangle + \lVert f \rVert^2 $$
whence $2\operatorname{Re}\langle h - f_0, f \rangle \le \lVert f \rVert^2$.
Write $\langle h - f_0, f \rangle = re^{i\theta}$. Replacing $f$ with $te^{i\theta}$ in the preceeding inequality, we have
$$ 2\operatorname{Re}\langle h - f_0, te^{i\theta} f \rangle = 2\operatorname{Re} \Big(te^{-i\theta} \langle h - f_0, f \rangle  \Big) = 2tr \le \lVert te^{i\theta} f\rVert^2 = t^2 \lVert f \rVert^2. $$
Taking $t \to 0$ we see that $r = 0$. I.e. $h - f_0 \perp f$.
For the converse, suppose $f \in \mathscr{M}$, then $f \perp h - f_0$ means that
$$ \lVert h - f \rVert^2 = \lVert (h - f_0) + (f_0 - f) \rVert^2 = \lVert h - f_0 \rVert^2 + \lVert f_0 - f \rVert^2 \ge \lVert h - f_0 \rVert^2. $$
Thus $\lVert h - f_0 \rVert = \operatorname{dist}(h, \mathscr{M})$.

Step 3: Note that this applies when $\mathscr{M}$ is finite dimensional, because finite dimensional subspaces are closed as a consequence of the Bolzano-Weirstrass Theorem.

