Showing that the WO closure of a *-algebra is a Von Neumann Algebra I think it's best to defer to the source that I'm reading for a statement of exactly what I need to prove.  Please refer to statement EP6 found on p.20 of this source.
The trouble is I don't follow any of the proof.  Literally.  When he says algebra, does he mean unital?  In the proof, he makes a claim about the join of two projections, then a claim about spectral theory, and then somehow these two come together to prove the theorem.  Please either show me another proof, or explain the meaning (unital or not is mostly the ambiguity for me) and proof of the first two claims, and then why these imply the theorem.  Thanks.
 A: EP6 addresses the situation where your (not necessarily unital) *-algebra $A$ is degenerate, i.e. of the form
$$
A=\begin{bmatrix}B&0\\0 & 0\end{bmatrix}.
$$
This is undesirable because when you take the strong closure (or double commutant) you get
$$
\overline{A}=\begin{bmatrix}\overline{B}&0\\ 0& 1\end{bmatrix}
$$
(with the undesired identity in 2,2 of the matrix). Here the $2\times 2$ matrices are written with respect with the decomposition $\mathcal{H}=\mathcal{K}\oplus\mathcal{W}$ in the notation from the notes. 
In the proof you want to see that $B$ is an algebra acting on $\mathcal{K}$. Note that
$$
\mathcal{K}=\mathcal{W}^\perp=\left(\bigcap_{a\in A}\ker(a)\right)^\perp=\overline{\bigcup_{a\in A}\ker(a)^\perp}.
$$
From this one deduces (by definition of the join of projections) that
$$
1_\mathcal{K}=\bigvee_{a\in A}P_{\ker(a)^\perp}.
$$
Now, $P_{\ker(a)^\perp}=P^\phantom{X}_{\text{ran}(a^*)}$. If you consider the case where $a=a^*$, then one can see by functional calculus that $P_{\text{ran}(a)}$ is in the closure of $A$ (one could also use the polar decomposition and do it for arbitrary $a$, but the selfadjoint will be enough). For this, one notices that $P_{\text{ran}(a)}=\chi_{\sigma(a)\setminus\{0\}}(a)$. The argument in the proof of EP5 shows that joins of projections are in the strong closure. And since for any $a$ we have $\ker a=\ker a^*a$, if we consider the kernels of selfadjoint elements we are already considering all kernels. Then $1_\mathcal{K}$ is in the closure of $A$.
So, we now know that we have a net $\{a_j\}\subset A$ with $a_j\to 1_\mathcal{K}$. But then $a_j1_\mathcal{K}\to1_\mathcal{K}$ too, and this shows that $1_\mathcal{K}$ is in the closure of $B$ when we see it inside $B(\mathcal{K})$.
