Approximate eigenvalue and continuous spectrum

So we assume that the range of $A-\lambda I$ is dense, and we can assume that $\lambda$ is not an eigenvalue (since otherwise the result is trivial). But then $A-\lambda I$ has trivial kernel, and it is not invertible -- because $\lambda\in\sigma(A)$ --, so we conclude that $A-\lambda I$ is not bounded below; this is exactly what you are looking for.

Since $\lambda \in \sigma (A)$, the operator $A-\lambda I$ has no bounded inverse. Hence $$\inf_{|v|=1} |Av-\lambda v|=0.$$ Otherwise, you could boundedly define $(A-\lambda I)^{-1}$ on the dense subspace $\operatorname{Im}(A-\lambda I)$, and $\lambda \in \rho (A)$.