Ideal of "Compact Operators" in a W*-algebra which gives the sigma-strong-* topology.

I don't believe that (3) characterises the $sigma$-strong* topology. Here's what I think is a counter-example.

Let $H=ell^2$. Consider $ell^infty$ acting on $H$ in the obvious way, so $c_0$ also acts on $H$ as compact operators. Let $K$ be the compact operator induced by $(1/sqrt k)_{kgeq 1}in c_0$.
I'm going to build a net $(x_beta)$ in $ell^inftysubseteqmathcal B(H)$ such that $x_betarightarrow 0$ $sigma$-strong* but with $|x_beta K|geq 1$ for all $beta$, so that $(x_beta)$ does not tend strictly to 0.

To do this, observe that the $sigma$-strong* topology, restricted to $ell^infty$, is given by seminorms of the form $x=(x_k) mapsto sum_k |x_k|^2 |b_k|$ where $(b_k)inell^1$. So, given $b^{(1)},cdots,b^{(n)}inell^1$, let $c_k = sum_{j=1}^n b^{(j)}_k$. Then $(c_k)inell^1$, and if $sum_k |x_k|^2 |c_k|$ is small, then certainly $sum_k |x_k|^2 |b^{(j)}_k|$ is small for each $j$.

So it suffices to show that for each $(c_k)inell^1$ and $epsilon>0$, we can find $xinell^infty$ with $sum_k |x_k|^2|c_k|<epsilon$, but with $|x_k|/sqrt kgeq 1$ for some $k$. We can do this by setting $x_k=0$ unless $k=N$ in which case $x_N = sqrt N$. This follows, as if we cannot do this, then $N|c_N|geqepsilon$ for all $N$, and so $sum_k |c_k| geq epsilonsum_k 1/k = infty$, a contradiction.

Did you mean to restrict to bounded sets?? Anyway, I wonder what happens for, say, $L^infty([0,1])$; I'd be surprised if this contained non-trivial ideals which were its own multiplier algebra.