"Crushing complement of open $n$-ball in an (compact) $n$-manifold to a point" is homeomorphic to $S^n$

Question

Let $M$ be an (compact?) $n$-manifold with $x\in M$. Then we have an open nbhd $U$ of $x$ such that $U$ is homeomorphic to an open $n$-ball $B$ in $\mathbb{R}^n$. be I am trying to prove the following:

$M/(M\setminus U)\cong S^n$.

I understand $D^n/\partial D^n\cong S^n$, and supposedly this is useful here, but I don't see how.