Let $(H, \langle \cdot, \cdot))$ be a real Hilbert space and $|\cdot|$ its induced norm. Let $(E_n)$ be a sequence of mutually orthogonal closed vector subspaces of $H$. This implies $\langle x, y \rangle=0$ for all $x \in E_m$ and $y\in E_n$ with $m \neq n$. Let $\pi_n:H \to H$ be the projection map onto $E_n$, i.e., $\pi_n (x)$ is the orthogonal projection of $x$ onto $E_n$. Let $E$ be the closure of $\operatorname{span}_\mathbb R (\bigcup_n E_n)$. Let $\pi:H \to H$ be the projection map onto $E$. Then $\pi, \pi_n$ are bounded linear operators. Let $(\mathcal L(H), \| \cdot\|)$ be the normed space of all bounded linear operators from $H$ to $H$. Let $\varphi_n := \sum_{k=1}^n \pi_k \in \mathcal L(H)$. I would like to prove that
$|\varphi_n (x) - \pi (x)| \xrightarrow{n \to \infty} 0$ for all $x \in H$.
Could you have a check on my below attempt?
Fix $x \in H$. Let's prove that $(\varphi_n (x))_n$ is convergent. Let $y_n := \pi_n (x)$. Then $\langle x-y_n, y_n \rangle =0$ and thus $\langle x, y_n \rangle= |y_n|^2$. Then $$ \langle x, \varphi_n (x) \rangle = \sum_{k=1}^n |y_k|^2 = |\varphi_n (x)|^2. $$
Then $|\varphi_n (x)| \le |x|$ for all $n$, so $\sum_{k=1}^\infty |y_k|^2 \le |x|^2 < \infty$. For $m <n$, we have $$ |\varphi_n (x) - \varphi_m (x)|^2 = \sum_{k=m}^{n-1} |y_k|^2. $$
It follows that $(\varphi_n (x))_n$ is a Cauchy sequence and thus convergent in $H$.
Let $v \in \bigcup_{k=1}^n E_k$. We assume that $v \in E_{\overline k}$ for some $\overline k \in \{1, 2, \ldots, n\}$. Then $$ \begin{align} \langle x - \varphi_n (x), v \rangle &= \left \langle x - \pi_{\overline k} (x) + \sum_{k=1, k \neq \overline k}^n \pi_k (x), v \right\rangle \\ &= \langle x - \pi_{\overline k} (x) , v \rangle + \sum_{k=1, k \neq \overline k}^n \langle \pi_k (x), v \rangle. \end{align} $$
Because $v \in E_{\overline k}$, we have $\langle x - \pi_{\overline k} (x) , v \rangle=0$. By mutual orthogonality of $(E_n)$, we get $\sum_{k=1, k \neq \overline k}^n \langle \pi_k (x), v \rangle =0$. This implies $$ \langle x - \varphi_n (x), v \rangle = 0 \quad \forall v \in E. $$
The claim then follows by taking the limit $n \to \infty$.