Removing finitely many points from a connected manifold (dimension greater then 2) remains connected?

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I think yes. Here is my approach.

As we know manifolds are locally Euclidean, Suppose we remove a point say $p$ from the manifold M then $p$ has a neighbourhood say U which is homeomorphic to an open subset V of $R^{n}$, let $\phi$ be the homeomorphism between them. Then, Consider any continuous map,

$ f: V-\phi(p) \rightarrow \{0,1\}$ where {0,1} has the discrete topology.

Then, it suffices to show that the map $f$ is constant.

Suppose $f$ is not constant then $\phi^{-1}(0)$ and $ \phi^{-1}(1)$ are not empty open sets in $V-\phi(p)$ more precisely $V-\phi(p)=\phi^{-1}(0) \cup \phi^{-1}(1)$. So, $\phi^{-1}(0)$ is a non trivial clopen set. Hence a contradiction.

And f is constant.

Now, $U$ is connected. I don't know how's to proceed and extend it to the whole manifold?

Thanks for any help!

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Hint: Things get much easier if you use that a manifold is connected if and only if it is path connected.