To prove Brouwer's fixed point theorem on a closed disk:
For any continuous function $f$ that maps a closed disk in $\mathbb R^2$ to itself, there exists a point $x_0$ in that disk, such that $f(x_0)=x_0$.
intuitive "proof": Take two points $y,z$ that lie on the opposite of each other on the boundary of the disk. Consider the curve $c_1$ that is the set of points on one side of the boundary, e.g. all points on the boundary of the disk that are to the "right" of $y$. Let $c_2$ be the set of points on the other side of the boundary of the disk (i.e. the complement of $c_1$ on the boundary). Since $f$ is continuous and so are $c_1,c_2$, $f(c_1)$ and $f(c_2)$ are also continuous curves on the disk.
Now, consider a homotopy between $c_1$ and $c_2$, that is, a continuous deformation $h(x,t)$ from $c_1$ to $c_2$. Then every point on the disk will have to at some point $t$ be in this deformation, and the curves $f(x(t))$, will at some point have to intersect the curve $x(t)$. This intersection is a fixed point.
Questions:
Can this proof be formalized, and is it generally correct?
Where is the point in the proof that is most "handwavy" and that needs to be formalized?
I noticed afterwards: I now recognize an obvious mistake: The last step in my argument, that this intersection is a fixed point, is of course wrong: the fact that the curves intersect does not mean that the intersection point on the curve $x(t)$ is "the same point" on the curve $f(x(t))$.
Is there a way to "save" this proof from the mistake I made?