[proofplan]
We prove countable additivity by reducing it to the assumed continuity from below. Given pairwise disjoint Borel sets, we form the finite partial unions $\Delta_N=\bigcup_{n=1}^{N}\Omega_n$, which increase to the full union $\Omega$. The projection multiplication rule gives orthogonality of the projections attached to disjoint sets, and the finite projection identities give $E(\Delta_N)=\sum_{n=1}^{N}E(\Omega_n)$. Applying the assumed strong convergence $E(\Delta_N)\to E(\Omega)$ then gives exactly strong countable additivity.
[/proofplan]
[step:Derive finite additivity for two disjoint Borel sets]
Let $A,B\in\mathcal B(K)$ be disjoint. Define $D:=A\cup B$. Since $A\subset D$, the projection multiplication rule gives
\begin{align*}
E(D)E(A)=E(D\cap A)=E(A).
\end{align*}
Also $B=D\cap A^c$, where $A^c:=K\setminus A$, so
\begin{align*}
E(B)=E(D\cap A^c)=E(D)E(A^c).
\end{align*}
Because $E(A)$ and $E(A^c)$ are complementary orthogonal projections under the finite projection-valued measure axioms, $E(A^c)=I_H-E(A)$. Hence
\begin{align*}
E(B)=E(D)(I_H-E(A))=E(D)-E(A).
\end{align*}
Therefore
\begin{align*}
E(A\cup B)=E(D)=E(A)+E(B).
\end{align*}
[guided]
We first isolate the finite additivity fact that will be used for partial unions. Let $A,B\in\mathcal B(K)$ be disjoint Borel sets, and define $D:=A\cup B$. Since $A\subset D$, we have $D\cap A=A$. Applying the multiplication axiom for the projection-valued set function gives
\begin{align*}
E(D)E(A)=E(D\cap A)=E(A).
\end{align*}
Now use the complement of $A$ in $K$. Define $A^c:=K\setminus A$. Since $A$ and $B$ are disjoint and $D=A\cup B$, we have $B=D\cap A^c$. Applying the multiplication axiom again gives
\begin{align*}
E(B)=E(D\cap A^c)=E(D)E(A^c).
\end{align*}
The finite projection-valued measure axioms give complementary projections on complements: $E(A^c)=I_H-E(A)$. Substituting this identity into the preceding display yields
\begin{align*}
E(B)=E(D)(I_H-E(A)).
\end{align*}
Distributing composition over addition in $\mathcal L(H)$ and using $E(D)E(A)=E(A)$, we get
\begin{align*}
E(B)=E(D)-E(A).
\end{align*}
Rearranging this equality gives
\begin{align*}
E(A\cup B)=E(D)=E(A)+E(B).
\end{align*}
Thus the projection identities imply finite additivity for two disjoint Borel sets.
[/guided]
[/step]
[step:Compute the projections of finite partial unions]
Let $(\Omega_n)_{n=1}^{\infty}$ be a pairwise disjoint sequence in $\mathcal B(K)$. For each $N\in\mathbb N$, define
\begin{align*}
\Delta_N:=\bigcup_{n=1}^{N}\Omega_n.
\end{align*}
We prove by induction on $N$ that
\begin{align*}
E(\Delta_N)=\sum_{n=1}^{N}E(\Omega_n)
\end{align*}
as operators on $H$.
For $N=1$, this is the identity $E(\Omega_1)=E(\Omega_1)$. Suppose the identity holds for some $N\in\mathbb N$. Since $\Delta_N$ is disjoint from $\Omega_{N+1}$, the two-set finite additivity proved above gives
\begin{align*}
E(\Delta_{N+1})=E(\Delta_N\cup\Omega_{N+1})=E(\Delta_N)+E(\Omega_{N+1}).
\end{align*}
Using the induction hypothesis,
\begin{align*}
E(\Delta_{N+1})=\sum_{n=1}^{N}E(\Omega_n)+E(\Omega_{N+1})=\sum_{n=1}^{N+1}E(\Omega_n).
\end{align*}
This proves the finite union formula for every $N\in\mathbb N$.
[/step]
[step:Apply continuity from below to the finite partial unions]
Define
\begin{align*}
\Omega:=\bigcup_{n=1}^{\infty}\Omega_n.
\end{align*}
The sequence $(\Delta_N)_{N=1}^{\infty}$ is increasing and satisfies
\begin{align*}
\bigcup_{N=1}^{\infty}\Delta_N=\Omega.
\end{align*}
By the assumed strong continuity from below, for every $x\in H$,
\begin{align*}
E(\Delta_N)x\to E(\Omega)x
\end{align*}
in the norm of $H$.
[/step]
[step:Identify the strong limit with the countable sum]
Let $x\in H$ be arbitrary. From the finite union formula, for every $N\in\mathbb N$,
\begin{align*}
E(\Delta_N)x=\sum_{n=1}^{N}E(\Omega_n)x.
\end{align*}
Combining this identity with the convergence from below gives
\begin{align*}
\sum_{n=1}^{N}E(\Omega_n)x\to E(\Omega)x
\end{align*}
in $H$. Since $x\in H$ was arbitrary, the partial sums $\sum_{n=1}^{N}E(\Omega_n)$ converge to $E(\Omega)$ in the strong operator topology. This is countable additivity of $E$ in the strong operator topology.
[/step]
