We have a nondecreasing sequence $(V_n)_{n\in\Bbb N}$ of closed subspaces of a Hilbert space $(H,|\cdot|)$ with an inner product $\cdot$ and the orthoprojectors $P_n$ onto $V_n$. We also have the closure $V$ of $\bigcup_{n\in\Bbb N} V_n$.

Let $D_n:=P_n-P_{n-1}$ for $n\in\Bbb N$, where $P_0:=0$. Take any $x\in H$. Then $$|x|^2\ge|P_n x|^2=\sum_{k=1}^n|D_k x|^2,$$ so that $\sum_{k\in\Bbb N}|D_k x|^2\le|x|^2<\infty$. So, for any natural $m$ and $n$ such that $m<n$ we have $$|P_n x-P_m x|^2=\sum_{k=m+1}^n|D_k x|^2\to0$$ as $m\to\infty$. So, there exists $$P_\infty x:=\lim_{n\to\infty}P_n x.$$ If is easy to see that $P_\infty$ is an orthoprojector. Also, clearly $$P_\infty H\subseteq V.$$

Also, $(P_\infty H)^\perp\subseteq V^\perp$. Indeed, otherwise there is some $h\in (P_\infty H)^\perp\setminus V^\perp$. So, $h\cdot v\ne0$ for some $v\in V$. So, $h\cdot v_n\ne0$ for some $n\in\Bbb N$ and some $v_n\in V_n$. But $h\cdot v_n=h\cdot(P_\infty v_n)=0$, since $h\in (P_\infty H)^\perp$. This contradiction proves that $(P_\infty H)^\perp\subseteq V^\perp$.

So, $$P_\infty H=V.$$ So, the sequence $(P_n)$ converges to the orthoprojector $P_\infty$ onto $V$ in the strong operator topology.

However, in general the sequence $(P_n)$ does not converge in the operator norm, because $\|P_{n+1}-P_n\|=1$ unless $V_{n+1}=V_n$.

We have a nondecreasing sequence $(V_n)_{n\in\Bbb N}$ of closed subspaces of a Hilbert space $(H,|\cdot|)$ with an inner product $\cdot$ and the orthoprojectors $P_n$ onto $V_n$. We also have the closure $V$ of $\bigcup_{n\in\Bbb N} V_n$.

Let $D_n:=P_n-P_{n-1}$ for $n\in\Bbb N$, where $P_0:=0$. Take any $x\in H$. Then $$|x|^2\ge|P_n x|^2=\sum_{k=1}^n|D_k x|^2,$$ so that $\sum_{k\in\Bbb N}|D_k x|^2\le|x|^2<\infty$. So, for any natural $m$ and $n$ such that $m<n$ we have $$|P_n x-P_m x|^2=\sum_{k=m+1}^n|D_k x|^2\to0$$ as $m\to\infty$. So, there exists $$P_\infty x:=\lim_{n\to\infty}P_n x.$$ If is easy to see that $P_\infty$ is an orthoprojector. Also, clearly $$P_\infty H\subseteq V.$$

Also, $(P_\infty H)^\perp\subseteq V^\perp$. Indeed, otherwise there is some $h\in (P_\infty H)^\perp\setminus V^\perp$. So, $h\cdot v\ne0$ for some $v\in V$. So, $h\cdot v_n\ne0$ for some $n\in\Bbb N$ and some $v_n\in V_n$. But $h\cdot v_n=h\cdot(P_\infty v_n)=0$, since $h\in (P_\infty H)^\perp$. This contradiction proves that $(P_\infty H)^\perp\subseteq V^\perp$.

So, $$P_\infty H=V.$$ So, the sequence $(P_n)$ converges to the orthoprojector $P_\infty$ onto $V$ in the strong operator topology.

However, in general the sequence $(P_n)$ does not converge in the operator norm, because $\|P_{n+1}-P_n\|=1$ unless $V_{n+1}=V_n$.