Let $\rho(z)$ denote the spectral radius of $R(z)$, where $R:U \to \text{Hom}(\mathcal{L},\mathcal{L})$ and $U$ is a closed subset of $\mathbb{C}$, here $\mathcal{L}$ can be understood as a Banach subspace of functions. Let $\lVert\,\cdot\,\rVert $ denote the operator norm in $\text{Hom}(\mathcal{L},\mathcal{L})$ then $R(z)=\sum_{n\ge 1}z^nR_n$ where $R_n\in \text{Hom}(\mathcal{L},\mathcal{L})$ and,
$$\rho(z)=\lim_{n\to\infty} \lVert R(z)^n \rVert^{1/n}$$
And by hypothesis the spectral radius is less than $1$, and as $U$ is compact the maximum will be reached.
$$\max_{z\in U} \rho(z)<1 \tag{1}\label{eq}$$
The step I didn't understand is this,
For every $z\in U$ there exists $n=n(z)$ such that $\lVert R(z)^n \rVert<1/2$. Since $z \mapsto \lVert R(z) \rVert$ is continuous in $U$, there is some $r=r(z)$ such that $\lVert R(w)^n \rVert<3/4$ for all $w\in B_r(z)$. Sinze $U$ is compact, there exists some $N$ such that for all $z\in U,\, \lVert R(z)^N \rVert <3/4$.
So, my doubt is in this choice of $N$. I've detailed a little more what was done. I believe this is the way, but I'm not sure as I couldn't finish the proof. This is my attempt to understand what was done:
Since $\max_{z\in U} \rho(z)<1$, we have that $\lVert R(z)^n \rVert\to 0$, therefore for every $z\in U$ exists $n_z$ (that depends on $z$), $\lVert R(z)^n \rVert<1/2$ for all $n\ge n_z$. I understood that now fixing the $z$ and $n_z$, there is $r$ such that $\lVert R(w)^{n_z} \rVert<3/4$ for all $w\in B_r(z)$.
As $U$ is compact, there is a finite subcover of these balls, ie,
$$U\subseteq \bigcup_{i\in\Lambda} B_{r_i}(z_i)\quad \text{where for all } w\in B_{r_i}(z_i),\, \lVert R(w)^{n_i} \rVert<3/4$$
and $\Lambda=\{1,2,\cdots,k\}$. First I had thought of taking $N$ as the maximum of these $n_i$, but I think taking the sum would be better. Let $N=n_1+\cdots + n_k$ and $w\in U$, so
$$\lVert R(w)^{N} \rVert \le \lVert R(w)^{n_1} \rVert\cdot \lVert R(w)^{n_2} \rVert\cdots \lVert R(w)^{n_k} \rVert $$
Without loss of generality we can assume that $w\in B_{r_1}(z_1)$, so
$$\lVert R(w)^{N} \rVert \le \lVert R(w)^{n_1} \rVert\cdot \lVert R(w)^{n_2} \rVert\cdots \lVert R(w)^{n_k} \rVert \le \frac{3}{4} \cdot \lVert R(w)^{n_2} \rVert\cdots \lVert R(w)^{n_k} \rVert$$
So what I'm having trouble with is how to ensure that the remainder is at least less than 1, to get the 3/4 of the proof. Because note that $n_i$ depends on $z$ and it might happen that $n(w)$ is greater than all $n_i$ so I can't guarantee that $\lVert R(w)^{n_i} \rVert<1/2$.
I don't know if this is the best choice for $N$, but I couldn't think of another one. Does anyone suggest something?
Another thing I thought was to see if it was possible to change the order of maximum and limit in \eqref{eq}, but for that I would need the sequence to be decreasing and I don't think I can guarantee that.
Edit: It seems to me that this is a standard result, maybe someone would also have a reference where to find it?
Reference: If reference is needed, this is the "preliminaries" of Lemma 4 of this article.
Your attempt is almost right. It only needs a small fix. In your attempt, you said:
"First I had thought of taking $N$ as the maximum of these $n_i$, but I think taking the sum would be better."
This statement is wrong. As you tried later, you can't make sure the remainder is less than 1 since it is not guaranteed that $\omega$ belongs to $B_{r_{n_i}}(z_i)$ for $i\geq 2$.
In fact, you only need to choose $N=\Pi_{i\in\Lambda}n_i$. Then for any $\omega\in U$, since $$ U\subseteq \bigcup_{i\in\Lambda} B_{r_i}(z_i)\quad \text{where for all } w\in B_{r_i}(z_i),\, \lVert R(w)^{n_i} \rVert<3/4\,, $$ there must exist $i^*$ such that $$ \omega\in B_{r_{i^*}}(z_{i^*})\,. $$ Then we have $$ \lVert R(\omega)^{N}\rVert\leq \left\| R(\omega)^{n^{i^*}}\right\|^{(N/n^{i^*})}\leq \frac{3}{4}\,. $$
Done