Show that the plane $\mathbb{R}^2 - \{(-1,0), (1,0)\}$ is homotopy equivalent to $C(1,0) \cup C(-1,0)$ I'm trying to find a deformation retract of the union of the two circles to $\mathbb{R}^2 - \{(-1,0), (1,0)\}$, any help is appreciated.
 A: If you understand a little geometry then you can write down the formulas you need. To do it with explicit care is intricate and somewhat tedious but straightforward.
I will use your notation $C(p)$ for the circle of radius $1$ centered on a point $p$:
$$C(p) = \{a \, | \, d(p,a) = 1\}
$$
Letting $p_1 = (-1,0)$ and $p_2=(1,0)$, you are asking for a deformation retraction
$$f :\mathbb{R}^2 - \{p_1,p_2\} \to C(p_1) \cup C(p_2)
$$
The value of $f(a)$ is defined in eight parts.
(1 and 2) For $i=1,2$, if $0 < d(p_i,a) \le 1$ then $f(a) = p_i + \frac{a-p_i}{|a-p_i|}$. This part of the function is just radial projection from inside of the circle $C(p_i)$ minus its center $p_i$ out to $C(p_i)$.
(3) If $a=(x,y)$ with $x \le -1$ and $d(p_1,a) \ge 1$ then $f(a) = p_1 + \frac{a-p_1}{|a-p_1|}$. This is radial projection from the left half of the outside of $C(p_1)$ back to $C(p_1)$.
(4) If $a=(x,y)$ with $x \ge +1$ and $d(p_2,a) \ge 1$ then a similar radial projection from the right half of the outside of $C(p_2)$ back to $C(p_2)$.
(5) If $a = (x,y)$ with $-1 \le x \le 0$, $y\ge 0$, and $d(p_1,a) \ge 1$ then $f(x,y) = (x,\sqrt{1-(x+1)^2})$. This is just vertical projection from the portion of the plane above the "1st quartercircle" of $C(p_1)$ down to that quartercircle. 
(6, 7, and 8) Similar vertical projections for $a=(x,y)$ with: $-1 \le x \le 0$, $y\le 0$, and $d(p_1,a) \ge 1$; and with $0 \le x \le 1$, $y\ge 0$, and $d(p_2,a) \ge 1$; and with $0 \le x \le 1$, $y\le 0$, and $d(p_2,a) \ge 1$. These map the points above and below the line segment between the circles' centers to their respective quartercircles.
A: It is always useful to think of the objects you are working with as clay. In your case, you have doubly-punctured plane. Imagine you could stick your fingers inside the holes and push outward, thus increasing the size of the holes. Now imagine you can take all the clay around those two holes and push it inward toward the center of the two holes you created. What are you left with? Exactly two disjoint circles $S_1$, provided you did your deformation without overlapping the two.
To explicitely write down a deformation retract, you want to smoothly deform the space as I mentioned. Are you able to do so?
