Definition
Define positivity in terms of: $$\omega\geq0:\iff\omega(X^*X)\geq0$$ (This way it serves best for the GNS-construction.)
Problem
Given a C*-algebra $1\notin\mathcal{A}$.
Consider a linear functional $\omega:\mathcal{A}\to\mathbb{C}$.
Then one has the equivalence: $$\omega\geq0\iff\|\omega\|=\lim\omega(E)$$
Attempt
Reduce the problem to positive elements: $$\|Z\|\leq1:\quad|\omega(Z)|\leq|\omega(X_+)|+|\omega(X_-)|+|\omega(Y_+)|+|\omega(Y_-)|$$ Remember the estimate on norms: $$Z=\sum_{\alpha=0\ldots3}i^\alpha Z_\alpha:\quad\|Z_\alpha\|\leq\|Z\|$$
Now, how to establish a bound: $$A\geq0:\quad\omega(A)\leq\|\omega\|_+<\infty\quad(\|A\|\leq1)$$ (I found it in Bratelli & Robinson. Unfortunately, I don't understand it.)
Proceed with the overall enclosing inequality: $$|\omega(A)|^2\leftarrow|\omega(AE)|^2\leq\omega(A^*A)\omega(E^2)\leq\|\omega\|_+^2\leq\|\omega\|^2$$ (Beware for limessuperior as the square is not operator-monotonic!)
Especially one has: $\lim\omega(E)=\sup\omega(E)=\|\omega\|_+=\|\omega\|$
Attention
Meanwhile, I got it. Answers are still heartly welcome!! :)
Disclaimer
My excuses; meanwhile I understood the proof.
Proof
Assume for contradiction that: $$A_n\geq0:\quad\omega(A_n)\geq n\quad(\|A_n\|\leq1)$$
Consider the absolutely convergent series: $$S_N:=\sum_{n=1}^N\frac{1}{n^2}A_n\to S\in\mathcal{A}$$ (Remember that the positive cone is closed!)
By positivity one has an upper bound: $$S_N\leq S\implies\omega(S_N)\leq S$$
Then one derives a contradiction: $$\infty\leftarrow\sum_n\frac{1}{n}\leq\omega(S_N)\leq\omega(S)$$
(Remember that increasing positive elements increases norm!)