In Evans' PDE book (p. 437) he defines the resolvent set of a closed linear operator, $A:D(A)\to A(D(A))$, on a Banach space, $X$, as follows:
i) We say a real number $\lambda$ belongs to $\rho(A)$, the $resolvent\text{ }set$ of $A$, provided the operator $\lambda I-A:D(A) \to X$ is one-to-one and onto.
My question is the following:
How do we know that the range of $\lambda I -A$ is $X$? My thought is that it has to do with the fact that $A$ is closed, but I am not sure how (we are not explicitly given that $D(A)$ is dense in $X$ either).
This is merely a definition, not a statement to be proved. The requirement for $\lambda$ to be in the resolvent set is that the operator $\lambda I-A$ maps $D(A)$ bijectively onto $X$. This is further emphasized by next item of the definition, where the resolvent operator $R_\lambda$ defined on all of $X$.
In infinite dimensions, it is quite possible for a linear operator to map a proper subspace onto the entire space. For example, the backward shift in $\ell^2$, $$ Sx = (x_2, x_3, x_4, \dots) $$ maps the proper subspace $\{x\in \ell^2 : x_1=0\}$ (which is not even dense) onto $\ell^2$.