I am witting concerning the example II.6 given in the paper "The General Stone-Weierstrass Problem" by C. A. Akemann:
Let me review this example. Let $H_i$ be the two dimensional Hilbert space $\mathbb{C}^2$ for $i=1,2,\cdots$ and consider the Hilbert space $H=\oplus H_i$. Let $A=\sum_1^{\infty} B(H_i)$ which is a C*-sub-algebra of $B(H)$. Two projection $p$ and $q$ are given in $A$ as follow:
For every $i\in \mathbb{N}$, let $p_i$ and $q_i$ be two projections in $B(H_i)$ such that: $p_i\vee q_i$ is the unit of $B(H_i)$ and $||p_i-q_i||\leq\frac{1}{2^i}$. (It is possible. For example, let $p_i=e_1\otimes e_1$ and $q_i= (\frac{1}{1+\frac{1}{4^i}})(e_1+\frac{1}{2^i}e_2)\otimes(e_1+\frac{1}{2^i}e_2)$.) We also put $p=\sum p_i$ and $q=\sum q_i$ where both $p,q$ are in $A$.
It is claimed that $p\vee q\ne1$ ( $1$ is the unit of $B(H)$). Equivalently, the range of the projection $p\vee q$ is not $H$. I do not get this point. Actually, I could not find any vector $\zeta\in H$ which is in the orthogonoal complement of $p\vee q(H)$.
That is not what Akemann claims. You avoided copying "(in $A^{**}$)" in your transcription; Akemann does not consider the $B(H)$ you defined above.
If, for instance, you let $\omega\in\beta\mathbb N$ be a free ultrafilter and let $f\in A^*$ be given by $f((x_j)_j)=\lim_\omega (x_j)_{22}$, then you get a state such that $f(p)=f(q)=0$. So, in $A^{**}$, $p\vee q$ belongs to a proper ideal.