How do you prove that the dunce cap is not a surface?

The dunce cap results from a triangle with edge word $aaa^{-1}$. At the edge, a small neighborhood is homeomorphic to three half-disks glued together along their diameters. How do you prove this is not homeomorphic to a single disk?

If you have two homotopic maps $f,g\colon S^1 \to X$, then $X \cup_f D^2$ is homotopy equivalent to $X \cup_g D^2$.

You can use this to show that the dunce cap is homotopy equivalent to $D^2$, and thus contractible. Since no closed surface is contractible (using classification of surfaces), the dunce cap is not a surface.

$D^2$ is the closed unit disk. By $X \cup_f D^2$, I mean gluing $D^2$ via the map $f\colon S^1 = \partial D^2 \to X$. This is the quotient space of $X \sqcup D^2$ identifying each point of $\partial D^2$ with its image under $f$ in $X$. So in our specific case, $D^2$ is homeomorphic to $S^1$ glued to $D^2$ under the identity map $S^1 \to S^1$. On the other hand, we have that the dunce cap is constructed by gluing $D^2$ to $S^1$ under the map $g\colon S^1 \to S^1$ given by $$ g(e^{i\theta}) = \begin{cases} \exp(4 i \theta) & 0 \leq \theta \leq \pi/2\\ \exp(4 i (2 \theta - \pi)) & \pi/2 \leq \theta \leq 3\pi/2\\ \exp(8 i(\pi - \theta)) & 3\pi/2 \leq \theta \leq 2\pi \end{cases}$$

It is not hard to show that $g$ is homotopic to to the identity map, and so (using the result I mentioned above), $D^2$ is homotopy equivalent to the dunce cap. So the dunce cap must be contractible.

Edit: I have now realized that the above answers the question in the title, which is not the question posed by the OP. To see that the dunce cap is not homeomorphic to $D^2$, you can simply note that the dunce cap is a disk glued along its boundary (albeit in a strange way), and thus has no 2-dimensional boundary, while $D^2$ does.