Let $A\in B(H)$ and $\{h_n\}$ be a dense sequence in the unit sphere of $H$.Suppose that there exists a sequence of mutually projections $\{P_i\}$. Denote $Q_n=\sum_{i\leq n}P_i$, if we assume that there exists $\delta＞0$ such that $\|Q_nh_n\|＞\delta$ fo all $n$.
Then $Q_n$ strongly increases to the identity operator $I$.
I tried to prove that for any $\epsilon＞0$, there exists $N$ when $n\geq N$, we have $\|Q_nh-h\|＜\epsilon$ for all $\|h\|\leq 1$. Since $\{h_n\}$ be a dense sequence in the unit sphere of $H$, for each $\epsilon＞0$, there exists $h_{n_0}$ such that $\|h-h_{n_0}\|＜\epsilon$. But how to use the condition "$\|Q_nh_n\|＞\delta$ fo all $n$" to show that $\|Q_nh-h\|＜\epsilon$ for all $\|h\|\leq 1$?
The sot limit $Q$ exists because the sequence $\{Q_n\}$ of projections is increasing. It is easy to show that a sot limit of projections is a projection, so $Q$ is a projection. Since $Q\geq Q_n$ for all $n$, we have $$ \|Qh_n\|^2=\langle Qh_n,Qh_n\rangle=\langle Qh_n,h_n\rangle\geq\langle Q_nh_n,h_n\rangle=\|Q_nh_n\|^2\geq\delta^2. $$
Suppose that $Q$ has kernel. That is, there exists $x$ in the unit sphere with $Qx=0$. By the density, there exists $h_n$ with $\|x-h_n\|<\delta/2$. Then $$ \delta<\|Qh_n\|=\|Qh_n-Qx\|=\|Q(h_n-x)\|\leq\|h_n-x\|\leq\frac\delta2, $$ a contradiction.