No, you just have to take $K_*$ such that $K \leq (\alpha_*-\alpha)\rho(x,y)$, and so for example $K_* = \frac{K}{\alpha-\alpha_*}$$K_* = \frac{K}{\alpha_*-\alpha}$.
If I am not missing something, a good way of doing the proof could be:
Since $\alpha<1$, there exists $\alpha_*\in (\alpha,1) \subset (0,1)$. Now take $K_* = \frac{K}{\alpha-\alpha_*}$$K_* = \frac{K}{\alpha_*-\alpha}$ and $\rho(x,y) ≥ K_*$. Then $$ \begin{align*} \alpha\,\rho(x,y) + K &≤ \alpha\,\rho(x,y) + (\alpha_*-\alpha)\,K_* \\ &≤ \alpha\,\rho(x,y) + (\alpha_*-\alpha)\,\rho(x,y) = \alpha_*\,\rho(x,y) \end{align*} $$$$ \begin{align*} \alpha\,\rho(x,y) + K &= \alpha\,\rho(x,y) + (\alpha_*-\alpha)\,K_* \\ &≤ \alpha\,\rho(x,y) + (\alpha_*-\alpha)\,\rho(x,y) = \alpha_*\,\rho(x,y) \end{align*} $$