Prove that removing an open 2-cell from $S^2$ results in a contractible space

Question

Let $X$ be a cellular decomposition of $S^2$. I want to show that if $r\in X^{(2)}$ then $X\setminus \text{Int}(r)$ is a contractible space. I don't know much topology so I don't know if this it completely trivial or requires a bit of work. I can see why it should be true but I cannot rigorously prove it.

Answer 1

Note that $S^2 \setminus \operatorname{Int}(r)$ is a deformation retract of $S^2 \setminus \{x\}$ where $x \in \operatorname{Int}(r)$ is any point, as a consequence of the fact that $D^2 \setminus \{y\}$ where $y$ is any interior point of $D^2$ deformation retracts onto the boundary $S^1 \subset D^2$. But $S^2 \setminus \{x\}$ is contractible since it is homeomorphic to $\mathbb{R}^2$, hence so is $X \setminus \operatorname{Int}(r)$.