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 $B(\mathcal{K})$.