Let $\mathbb{B}(\mathcal{H})$ denote the $C^*$-algebra of all bounded linear operators on a complex Hilbert space $\mathcal{H}$ endowed with an inner product $(.\mid.)$.
Let $T_1,T_2\in\mathbb{B}(\mathcal{H})$. Suppose that there exist a sequence of unit vectors $\{\xi_n\}$ in $\mathcal{H}$ and $\lambda\in\mathbb{T}$ (i.e. $|\lambda|=1$) such that $\lim_{n\rightarrow\infty} (T_1\xi_n\mid T_2\xi_n)=\lambda\|T_1\|\,\|T_2\|$. It follows from $$\|T_1\|\,\|T_2\|=\lim_{n\rightarrow\infty} |(T_1\xi_n\mid T_2\xi_n)|\leq\lim_{n\rightarrow\infty} \|T_1\xi_n\|\,\|T_2\|\leq\|T_1\|\,\|T_2\|,$$ that $\lim_{n\rightarrow\infty} \|T_1\xi_n\|=\|T_1\|$ and by using a similar argument, $\lim_{n\rightarrow\infty} \|T_2\xi_n\|=\|T_2\|$. So that $$\lim_{n\rightarrow\infty} \mbox{Re}(T_1\xi_n\mid \lambda T_2\xi_n)=\lim_{n\rightarrow\infty} (T_1\xi_n\mid \lambda T_2\xi_n)=\|T_1\|\,\|T_2\|.$$
I don't understand the last line. Why $$\lim_{n\rightarrow\infty} \mbox{Re}(T_1\xi_n\mid \lambda T_2\xi_n)=\lim_{n\rightarrow\infty} (T_1\xi_n\mid \lambda T_2\xi_n)=\|T_1\|\,\|T_2\|?$$
Read it from right to left. From the initial assumption $$\lim_{n \to \infty}\: (T_1\xi_n \mid T_2 \xi_n) = \lambda \lVert T_1\rVert \lVert T_2\rVert$$ we directly obtain $$\lim_{n \to \infty} \: (T_1\xi_n \mid \lambda T_2\xi_n) = \lVert T_1\rVert \lVert T_2\rVert$$ since the inner product is conjugate-linear in the second argument and $\lvert\lambda\rvert = 1$. And now, consider an arbitrary convergent sequence $(z_n)$ of complex numbers. Then by continuity of the real part $$\lim_{n \to \infty} \operatorname{Re} z_n = \operatorname{Re} \lim_{n \to \infty} z_n\,.$$ If, as in our case $z_n = (T_1\xi_n \mid \lambda T_2\xi_n)$, the limit is real we obtain $$\lim_{n \to \infty} \operatorname{Re} z_n = \lim_{n \to \infty} z_n\,.$$