How to show weak*-closed in the unit ball? I have a unital C* algebra $\mathcal A$ and a subset $K \subseteq B_1^*  \subseteq \mathcal A^*$ of the unit ball in the dual space. I want to show that K is compact in order to apply the Krein-Milman Theorem, which requires $K$ to be so.
I am stuck with two questions:
1) Does "compact" here make sense in any other topology but the weak*-topology?
2) If not, I thought about using the Banach-Alaoglu theorem, by which it would suffice to show that $K$ is weak*-closed in $B_1^*$. How does one technically show weak*-closed?
EDIT:
To be specific, $K:= \{\omega \text{ positive}, \vert\vert \omega \vert\vert_{\mathcal A^*}=1, \omega(A^*A)=\vert\vert A \vert\vert^2\}$
 A: 
As explained by Ben Passer, we expect $K$ to be weak*-compact. But I think it should be noted that $K$ is empty when $\dim A\geq 2$. And when $A=\mathbb{C}$, $K$ is a singleton consisting of the state $\omega(\lambda)=\lambda$.

Indeed, if $\omega\in K$, then for every $h$ self-adjoint positive, we have $h=k^*k$ whence
$$
\omega(h)=\omega(k^*k)=\|k\|^2=\|k^*k\|=\|h\|.
$$
Now for every $h$ self-adjoint positive with $\|h\|=1$, we have $1-h$ self-adjoint positive whence
$$
\|1-h\|=\omega(1-h)=\omega(1)-\omega(h)=1-\|h\|=0\quad\Rightarrow \quad h=1
$$
where we used the key property of positive linear functionals on unital $C^*$-algebras that $\omega(1)=\|\omega\|$. So $\omega(1)=1$ when $\|\omega\|=1$.
It follows that for every self-adjoint positive element, $h=\|h\|1$, i.e. scalar. Then for every self-adjoint $k$, $h=\|k\|1-k$ is self-adjoint positive, whence $h$, and therefore $k$ is scalar. Finally, every element $a\in A$ is a linear combination of self-adjoint elements by
$$
a=\frac{a+a^*}{2}+i\frac{a-a^*}{2i}=k_1+ik_2\quad k_j^*=k_j
$$
so $a$ is scalar. That is $A=\mathbb{C}1$ is one-dimensional.
Now when $A=\mathbb{C}$, it follows from the assumption $\omega(t)=t$ for every $t\geq 0$, whence for every $t\in\mathbb{R}$, and finally over $\mathbb{C}$ by linearity.
