Why doesn't the circle retract to a point?

First of all, let's clarify one thing. A circle does retract onto a point, because a retract of a circle to a point on it is just a constant map $r : S^1 \to \{p\}$. What you're really asking about is the fact that a circle doesn't deformation retract onto a point.

A deformation retract would be a homotopy $F : S^1 \times I \to S^1$ taking the circle to one of its points, so to deformation retract a circle to a point you'd need to retract it to a point on the circle, via a series of maps $F_t$ that map the circle to itself. Your map seems to be shrinking the circle to a point, which doesn't work because you're moving the rest of the circle off of itself into the "empty space inside of it," which isn't allowed.

In other words, you're viewing the circle as being embedded in the plane, like you'd draw a circle on a piece of paper. But, topologically, the points "inside the circle" don't exist -- there's only the circle itself.

If I understand correctly, you are asking why a circle does not deformation retract onto a point. Recall that a deformation retract of $S^1$ onto a point $x_0\in S$ is a function $f:S^1\times I\to S^1$ such that $f(x,0)=x$ for all $x\in S^1$ and $f(x,1)=x_0$ for all $x\in S^1$. You propose drawing a cone with one end (corresponding to $0$) equal to $S^1$ and the other (corresponding to $1$) equal to $x_0$, and defining $f(x,t)$ to be the point obtained by travelling a distance $t$ along the line segment connecting $x$ at one end to $x_0$ at the other. The issue is that for $t\ne 0,1$ and $x\ne x_0$, we will have $f(x,t)\notin S^1$, so this map does not actually have range in $S^1$. Essentially, the problem with your cone visualization is that part of the cone lies outside $S^1$.