Lemma If $\|.\|_{\alpha}$ is a C*-norm on $\mathcal{A}\odot\mathcal{B}$, then it is a cross norm.
Thus, for a unit vector $\xi\in ef\mathcal{H}$, we have $\|(a\bigotimes b)(\xi)-\xi\|\le 2\varepsilon$. Since $\varepsilon$ was arbitrary, $\|a\bigotimes b\|=1$.
how to understand $\|(a\bigotimes b)(\xi)-\xi\|\le 2\varepsilon$. please help me!!!
As far as I can tell, a few details need adjustments. For instance, it should be $$e=\chi_{[(1-\varepsilon)^{1/2},1]}(a\otimes 1),\ \ \ \ \ \ \chi_{[(1-\varepsilon)^{1/2},1]}(1\otimes b).$$
You have, since everything commutes, $$ f(1\otimes b)f\geq(1-\varepsilon)^{1/2}f,\ \ \ e(a\otimes1)e\geq(1-\varepsilon)^{1/2}e. $$ Then \begin{align} fef-fe(a\otimes b)ef&=fef-(a\otimes1)^{1/2}ef(1\otimes b)fe(a\otimes1)^{1/2}\\ \ \\ &\leq fef-(a\otimes1)^{1/2}e(1-\varepsilon)^{1/2}f e(a\otimes1)^{1/2}\\ \ \\ &=fef-(1-\varepsilon)^{1/2}fe(a\otimes1)ef\\ \ \\ &\leq fef-(1-\varepsilon)fef\\ \ \\ &=\varepsilon \,fef. \end{align} Then $$ \|\xi-(a\otimes b)\xi\|=\|(fef-fe(a\otimes b)ef)\xi\|\leq\varepsilon. $$