Asking for clarification on a step for the proof of the inverse function theorem

This is from the book called Calculus and Analysis In Euclidean Space by Shurman et al., pp. 207

Theorem: Let $A$ be an open subset of $\mathbb{R}^n$, and let $f: A \to \mathbb{R}^n$ have continuous partial derivatives at every point of $A$. Let $a$ be a point of $A$, and suppose that $\det f'(a) \neq 0$. Then there exists an open set $V \subset A$ containing $a$ and an open set $W \subset \mathbb{R}^n$ containing $f(a)$ such that $f:V \to W$ has continuously differentiable inverse $f^{-1}:W \to V$. For each $y = f(x) \in W$, the derivative of the inverse is the inverse of the derivative $D(f^{-1})_{y} = (Df_x)^{-1}$.

The author first shows how we can assume without the loss of generality that $f'(a) = I_n$ (the identity matrix). After that the author begins to construct a closed ball $\overline{B}$, in order to apply the difference magnification lemma.

After that the author writes:

As $x$ varies continuously near $a$, the $(i, j)$th entry of $D_jf_i(x)$ of $f'(x)$ varies continuously near $\delta_{ij}$, and so the scalar $\det f'(x)$ varies continuously near $1$.

Question: I'm asking for clarification on the following step made in the proof.

Since $D_jf_i(x) - \delta_{ij} = 0$ and since $\det f'(a) = 1$, applying the persistence of inequality principle $n^2 + 1$ times shows that there exists a closed ball $\overline{B}$ about $a$ small enough that $$\left|D_jf_i(x) - \delta_{ij}\right| < \frac{1}{2n^2} \text{ for all }i, j \in {1,2,\dots,n}, x \in \overline{B}$$ and $\det f'(x) \neq 0, \text{ for all } x \in \overline{B}$

What I don't understand is i.) why does the persistence of the inequality principle show that there exists a closed ball with a radius of at least $\frac{1}{2n^2}$, ii.) why do we apply the principle $n^2 + 1$ times?

I get that the $\det f'(x) \neq 0$ for all $x \in \overline{B}$ results directly from the said principle, but given that the author has stated the principle as

Let $A$ be a subset of $\mathbb{R}^n$ and let $f:A \to \mathbb{R}^n$ be a continuous mapping. Let $p \in A$ and suppose $b \in \mathbb{R}^n \setminus \{f(p)\}$. Then there exists some $\epsilon > 0$ such that $\text{ for all } x \in A \text{ such that } |x - p| < \epsilon, f(x) \neq b$.

I don't immediately see how we construct the closed ball with the said radius.

I think one should use the very definition of continuity rather than the property the author stated. Here, we know that for each of the $n^2$ couples $(i,j)\in\{1,\dots,n\}^2$, the following holds: $$\forall\varepsilon>0,\exists\eta_{ij}>0\text{ s.t. }|x-a|<\eta_{ij}\implies|D_jf_i(x)-\delta_{ij}-0|<\varepsilon.$$ Get such $\eta_{ij}$ for all $(i,j)\in\{1,\dots,n\}^2$ for fixed $\varepsilon=\frac{1}{2n^2}$, then set $\eta=\min_{(i,j)\in\{1,\dots,n\}^2}\eta_{ij}$ (note we still have $\eta>0$). Then apply the persistence of inequality principle to $\det f'(a)\neq 0$, then take the minimum between the radius it gives you and $\eta$ to conclude.