Closedness of $R(\lambda -T)$ when $T$ closed symmetric operator

Question

Let $T$ be a closed symmetric operator defined on a Hilbert space.

I want to proof that the Rang $R(\lambda - T)$ is closed if and only if $Im(\lambda)\neq 0$.

I used the sequences technic to proof that if $Im(\lambda)\neq 0$ then the equality hold, but I couldn't see why it is not true if $Im(\lambda)= 0$ !!!

I proceed like the following

if $T$ is closed operator so $(\lambda - T)$ too, then $R(\lambda - T)$ is close.

Could someone give me a counter-example or any clarification, please?