Let $X$ be a topological space, let $Y=X \times [0,1]$ and $A=X \times \left \{ 0 \right \} \subset Y$. the cone on X is the space $\left \{ X \times [0,1] \right \}/\left \{ X \times \left \{ 0 \right \} \right \}$.
Now let $X=S^1$, my goal is to prove that the cone on $X$ is homeomorphic to $D^2$, where $D^2$ is the unitary closed disk in $\Bbb R^2$. Basically, this cone is obtained from a cylinder by contraction of the lower base to a point.
Using the polar coordinates we can establish a map $f: \left \{ X \times [0,1] \right \}/\left \{ X \times \left \{ 0 \right \} \right \} \rightarrow D^2$ defined as $f(\theta, \rho)= \rho(\cos \theta, \sin \theta)$.
This map is continuos, injective and surjective, but how to prove that is a homeomorphism?
To show $f$ is a homeomorphism without using compactness, you can explicitly construct the inverse of $f$, namely $$ g(x, y) = [t(x, y), \sqrt{x^2 + y^2}] $$ where brackets denote equivalence class, and $$ t(x, y) = \begin{cases} 0 & (x, y) = (0,0) \\ \frac{\pi}{2} & x = 0, y > 0\\ \frac{3\pi}{2} & x = 0, y < 0\\ \arctan(y/x) & x > 0 \\ \pi + \arctan(y/x) & x < 0 \end{cases} $$ where each of these values is to be considered a number (mod $2\pi$), hence an element of $S^1$.
The continuity of $g$ is pretty clear everywhere except along the $y$-axis. For those, you'll need to show that $g^{-1}(U)$ is open whenever $U$ is open in the domain, which will mean writing down a way to describe open sets in the quotient, which is a pain in the neck, but you're welcome to do it.