Following is from my lecture notes. I have resolvent map $R(\cdot,x): \mathbb{C} \rightarrow A$, where $x\in A$ Banach algebra. Domain is whole complex numbers because for the sake of contradiction it was assumed spectrum is empty.
Then one argues from $ \Vert R(\lambda,x) \Vert \leq \frac{1}{|\lambda | - \Vert x \Vert}$ for any lambda $|\lambda | > \Vert x \Vert$ that $R(\cdot,x)$ is bounded function on $\mathbb{C}$.
Question: Why is $R(\cdot,x)$ bounded?
When $|\lambda|\leq\|x\|$, as you say in the comments, $$ R(\lambda,x)=(\lambda-x)^{-1}=\sum_{k=1}^\infty\lambda^{-k}\,x^{k-1}. $$ Then $$ \|R(\lambda, x)\|\leq\sum_{k=1}^\infty|\lambda|^{-k}\,\|x\|^{k-1}=\frac1{|\lambda|-\|x\|}. $$ This requires using that $\Big\|\sum_k y^k\Big\|\leq\sum_k\|y\|^k$. This follows from the fact that the norm is continuous, the triangle inequality, and the submultiplicativity of the norm.
Here is another argument without the Neumann series. Given $a\in A$ such that $\|a\|<1$, then $1+a$ is invertible and $\def\abajo{\\[0.2cm]}$ \begin{align*} 1&=\|(1+a)(1+a)^{-1}\|=\|(1+a)^{-1}+a(1+a)^{-1}\|\abajo &\geq\|(1+a)^{-1}\|-\|a(1+a)^{-1}\|\abajo &\geq\|(1+a)^{-1}\|-\|a\|\,\|(1+a)^{-1}\|. \end{align*} Thus $$\tag1 \|(1+a)^{-1}\|\leq\frac1{1-\|a\|}. $$ Now, when $\|x\|<|\lambda|$ we apply $(1)$ to $a=x/|\lambda|$ to get $$ \|(\lambda-x)^{-1}\|\leq\frac1{|\lambda|-\|x\|} $$