For this problem, let $f$ be a mapping defined on a neighborhood of $0$ in $\mathbb{R}^n$ to $\mathbb{R}^n$ such that $f(0) = 0$. Denote by $\overline{B_r}$ the closed ball of radius $r$ centered at the origin. Assume that there exists a $\delta > 0$ such that $f(x) - x$ is Lipschitz continuous on $\overline{B_{\delta}}$ with Lipschitz constant less than $\frac{1}{2}$. That is, $\lVert f(x) - x - f(y) + y\rVert \leq c\lVert x - y\rVert$ for some $c < \frac{1}{2}$ for all $x,y$ in $\overline{B_{\delta}}$.
Show that $f$ is injective on $\overline{B_{\delta}}$
Prove that the image under $f$ of $\overline{B_{\delta}}$ contains $\overline{B_{\frac{\delta}{2}}}$ (Hint: show that for $y\in\overline{B_{\frac{\delta}{2}}}$, the map $g_y(x) = y + x - f(x)$ takes $\overline{B_{\delta}}$ to itself and has a fixed point.)
Conclude that $f$ has a continuous inverse on some domain containing $\overline{B_{\frac{\delta}{2}}}$
I have shown 1 and 2 but I can't figure out 3.
By 2. we have $\overline{B_{\frac{\delta}{2}}} \subset f(\overline{B_{\delta}})$. Then by 1. $f:\overline{B_{\delta}} \to f(\overline{B_{\delta}})$ is bijectice. So there is $f^{-1}:f(\overline{B_{\delta}}) \to \overline{B_{\delta}}$ inverse of $f$. Now, for every $x,y \in \overline{B_{\delta}}$ $$ ||f(x)-f(y)||=||f(x)-x-f(y)+y + (x-y)|| \leq ||f(x)-x-f(y)+y|| + ||x-y||. $$ Then $||f(x)-f(y)||\leq (c+1)||x-y||$ for all $x,y \in \overline{B_{\delta}}$. So $f:\overline{B_{\delta}} \to f(\overline{B_{\delta}})$ is continuous. But a continuous invertible function with compact domain has continuous inverse(Try to check this). Since $\overline{B_{\delta}}$ is compact we have that $f$ has a continuous invertible function on $f(\overline{B_{\delta}})$ that contains $\overline{B_{\frac{\delta}{2}}}$.