Q:

Prove that $\exists x_0 \in D$ such that $f(x_0) = (0,0)$

Here's the question (from last quarter's final):

Define $D$ to be the closed unit disk, that is $D = \{(x_1,x_2):x_1^2 + x_2^2 \leq 1\}$. Let $f:D \to \mathbb{R}^2$ be a $C^1$ mapping.  If $f'(x)$ is invertible for all $x \in D$ and
  $$
|f(x) - x| \leq \frac 13, \quad x \in D
$$
  prove that there exists a point $x_0 \in D$ such that $f(x_0) = (0,0)$.

Thoughts so far: clearly, the inverse function theorem applies, so that $f^{-1}$ can be defined in the neighborhood of any point.  Furthermore, the inequality applies so that
$$
|x - f^{-1}(x)| \leq \frac 13, \quad x \in f(D) 
$$
I'm not sure how any of this would allow me to show that $0 \in f(D)$, so I'm pretty much stuck for now.
Any input, be it a gentle nudge or full solution, would be very much appreciated.
EDIT: As copper.hat points out, the Brouwer fixed-point theorem makes quick work of this problem.  However, I am still looking for a solution "in the spirit of the question", hopefully one that uses the existence of an inverse in  the neighborhood of each point.

A:

Let $\phi(x) = x -f(x)$. Then $\|\phi(x)\| \le {1 \over 3} $ for all $x$ in the closed unit ball. Since the closed unit ball is compact and convex, the Brouwer fixed point theorem shows that $\phi$ has a fixed point in the unit disk. Hence $f(x) = 0$ at this fixed point.
Note: $\phi(x) = x$ iff $f(x) = 0$.
Alternative proof:
Let $\eta(x) = {1 \over 2 } \|f(x)\|^2$. Since $\eta$ is continuous, it has a minimum at some point $\hat{x} \in D$.
The hypotheses show that $\eta(x) \ge {2 \over 3} $ for $\|x\| = 1$, and $\eta(0) \le {1 \over 3}$, hence we have $\hat{x} \in D^\circ$, and so
${\partial \eta(\hat{x}) \over  \partial x} = 0$ at the minimum. Since
${\partial \eta(\hat{x}) \over  \partial x} = {\partial f(\hat{x}) \over  \partial x} f(\hat{x})$ and the derivative is  invertible, we have $f(\hat{x}) = 0$ as required.

