Neighborhood of inclusion in space of Lipschitz maps is 1-1

Let $B \subset \mathbb{R}^n$ be the closed unit ball. Let $i(x) = x$ denote the inclusion map. Let $\|\cdot\|$ be any norm. Given $f:B\to \mathbb{R}^n$, define the sup norm $\|f\|_\infty:=\sup_{x \in B} \|f(x)\|$.

Claim: There is a neighborhood $W$ of $i$ in the space of Lipschitz maps endowed with the sup-norm topology such that $\forall f \in W$: $f$ is $1$-$1$.

Motivation: I'm studying the paper "Persistence and Smoothness of Invariant Manifolds for Flows" by Neil Fenichel. This is part of the (unproven) "Implicit Function Theorem (for Lipschitz maps)" on page 207 of that paper -- he cites a textbook, but I couldn't find this result in the book he cites. Based on his hints that the proof is similar to the standard proof of either the smooth implicit or inverse function theorems, I believe I need to show an inequality of the form $\|f(x)-f(y)\| \geq C\|x-y\|$ for some $C > 0$.

Where I'm stuck: Given two Lipschitz functions $f$ and $g$, if I could prove that $\|f-g\|_\infty$ being small implies that the Lipschitz constant of $(f-g)$ is small, I think I can prove everything else. I've been trying to modify the proof on page 179 of "Advanced Multivariable Calculus" by Edwards to apply to Lipschitz functions with no luck so far.

Edit 1: The author only says "Lipschitz topology", so I am only guessing when I said "sup-norm" topology.

Edit 2: I've discovered the claim is not true as I stated it. I posted a counterexample below, and also formulated and proved an alternative statement.

Counterexample: I've proven that the claim is not true as I stated it. Here is a counterexample.

Let $i:B \to \mathbb{R}$ be the inclusion. Consider the family of functions $f_\varepsilon:B \to \mathbb{R}$ defined by $$f_\varepsilon(x):= x + \varepsilon \sin\left(\frac{2\pi}{\varepsilon^2}x\right).$$

Then $\|f_\varepsilon - i\| = \varepsilon \to 0$ as $\varepsilon \to 0$, but for sufficiently small $\epsilon > 0$ $f$ is always zero at $x = 0$ and also some positive value since the period of $f_\varepsilon$ is $\varepsilon^2$.

True statement:

I've proven this other result instead, which I think is what Fenichel intended.

Given any norm $\|\cdot\|$ on $\mathbb{R}^n$, define the Lipschitz norm on the space of Lipschitz functions $B \to \mathbb{R}^n$ by $$\|f\|_L:= \sup_{x,y \in B; x \not = y}\frac{\|f(x)-f(y)\|}{\|x-y\|}$$.

Then for any $0 < \varepsilon < 1$, $\|f-i\|_L < \varepsilon$ implies that $f:B \to \mathbb{R}^n$ is injective.

Proof: By assumption, $\forall x,y \in B$ we have $\|f(x)-i(x) - (f(y)-i(y))\| < \epsilon \|x-y\|$. Hence $\left|\|f(x)-f(y)\| - \|i(x)-i(y)\|\right| < \epsilon\|x-y\|$, so $$\|f(x) - f(y)\| > \|i(x)-i(y)\| - \epsilon \|x-y\| = (1-\epsilon)\|x-y\|.$$