- WLOG, assume \(p_0, p_1\) are the north and south poles of \(S^2\). We can then form a deformation retract of \(X\) onto the equator of \(S^2\), which is equal to \(S^1\). To do so, just move every point \(x\) along the unique great circle connecting \(x, p_0, p_1\), and proceed at linear speed towards the equator. This is well defined at every point on \(S^2\) except the poles, which are not included in \(X\), and the equator is fixed at every instant. So this forms a deformation retract. Alternatively, use the fact that \(\RR^n -\pt \cong S^{n-1} \cross \RR\) via polar coordinates, and \(S^n - \pt \cong \RR^n\) by stereographic projection. So \(S^2 - \theset{p_0, p_1} \cong \RR^2 - \theset{p_1} \cong S^{1} \cross \RR\). But since \(\RR\) is contractible, the last one is homotopic to \(S^1 \cross \theset {0} \cong S^1\). Alternatively: use the lemma, then \(k=2\) and so \(S^2 - \theset{p_1, p_2} \homotopic \bigvee_{i=1}^{1}S^1 = S^1\).
Solution
1. Removing one point from \(S^2\) gives \(\mathbb R^2\) by stereographic projection; the second removed point becomes one point of \(\mathbb R^2\).
Choose one of the two deleted points as the projection point.
2. The punctured plane \(\mathbb R^2\setminus\{0\}\) strongly deformation retracts onto the unit circle by\[H(x,t)=\bigl((1-t)+t/\|x\|\bigr)x.\]
For \(x\ne0\) the scalar is positive, so the homotopy stays in the punctured plane; at \(t=1\) it is radial projection and it fixes the unit circle.
3. Therefore\[\boxed{S^2\setminus\{p,q\}\simeq S^1.}\]
Combine <1>1–<1>2.