STANDING ASSUMPTIONS: Let $T:D_T\rightarrow X$ be a linear operator, where $X$ is a normed space (or metrizable TVS) and $D_T\subset X$.

DEFINITION: Then $\lambda$ belongs to the *resolvent set* $\rho(T)$ iff $T-\lambda I$ is one-to-one and **onto** and its inverse is bounded.

LEMMA. **A linear operator $T$ in a normed space $X$ is closed if$^\dagger$ it has a non-empty resolvent set.**

PROOF. Let $D_T\owns x_n\rightarrow x$, $Tx_n\rightarrow y$.$^\ddagger$ Let $\lambda$ be in the resolvent set. Then $x=\lim_n x_n = \lim_n (T-\lambda)^{-1}(T-\lambda)x_n=(T-\lambda)^{-1}(y-\lambda x)$ (+).
Therefore, $x\in (T-\lambda)^{-1}(X)=D_T$.
By (+), $(T-\lambda)x=(T-\lambda)(T-\lambda)^{-1}(y-\lambda x)=(y-\lambda x)$;
hence $Tx=y$, so $T$ is closed, QED.

Note: Robert Israel's shorter proof on this page shows the same for any TVS (if you require that the inverse is a continuous operator). The above proof (that works for all metrizable TVSs) might be easier for some readers.

$\dagger$) The "if" in the lemma cannot be reversed, as some closed operators (even on $\ell^2$) have an empty resolvent set, by https://math.stackexchange.com/questions/3262168/closed-operator-with-trivial-resolvent-set

$\ddagger$) In a metrizable space, a set is closed iff it is sequentially closed. So the graph $G_T:=\{(x,Tx):\, x\in D_T\}$ is closed iff $G_T\owns (x_n,Tx_n)\rightarrow (x,y)\ \Rightarrow\ (x,y)\in G_T$ (i.e., $x\in D_T$ and $y=Tx$).

Note that your assumption that the operator is densely defined was not needed. This is a strict improvement:

EXAMPLE. Let $L$ denote the inverse of the right-shift $R:(x_1,x_2,\cdots)\rightarrow (0,x_1,x_2,\cdots)$. Then dom$L=R(\ell^2)$ is not dense, as $(1,0,0,\cdots)$ is not in its closure, but yet the (onto-)resolvent set $\rho(L)$ is nonempty, as $(L-0)^{-1}=R$ and hence $0\in\rho(L)$.

This "onto" definition seems to be standard:
Rudin: F.A., 13.26 &
https://en.wikipedia.org/wiki/Spectrum_(functional_analysis)#Spectrum_of_an_unbounded_operator

**Your claim is true only for this standard definition ("onto"), not for the older definition ("dense range" in place of "onto")**, by Matthew Daws' example. For closed operators, the two definitions are equivalent. For a further discussion on this and the two definitions, see: Standard definition of a resolvent: A-zI must be onto, not merely have a dense range?