Let $M$ be a smooth manifold, of dimension $m$. Take a point $p \in M$ and define the new manifold $M'=M-\{p\}$, which does not depend on the point chosen up to diffeomorphism ,because manifolds are homogeneous.
Now take an embedding $f: D^m \hookrightarrow M$such that $f(0)=p$ and take the smooth manifold $M''=M-f(D^m)$.This manifold does not depend up to diffeomorphism on the chosen embedding, because of the disk theorem.
I would like to show that $M''$ is diffeomorphic to $M'$, but I do not really know how to do that.
Choose a diffeomorphism $\phi: D^n_{1} \setminus \{0\} \to D^n_1 \setminus D^n_{1/2}$, which is the identity in a neighborhood of the boundary; here the subscripts denote radius of the disc. Such a diffeomorphism may be induced by a diffeomorphism $(0,1] \cong (1/2, 1]$ which is the identity near the boundary.
Now pick a chart in $M$ around $p$ (identifying an open subset $U \subset M$ with $\Bbb R^n$, sending $p$ to $0$; let your disc $D$ be the subset of $U$ corresponding to $D^n_{1/2} \subset \Bbb R^n$). On $D^n_1 \subset \Bbb R^n \cong U$, we have already defined a diffeomorphism $\phi: D^n_1 \setminus \{0\} \to D^n_1 \setminus D^n_{1/2}$. Extending by the identity to the rest of $\Bbb R^n$, and passing instead to $U$, this extends to a diffeomorphism $U \setminus \{p\} \to U \setminus D$.
You now extend by the identity to the rest of $M \setminus \{p\} \to M \setminus D$.