Existence of Smooth path in a Domain through a Sequence of Points The following question seems intuitively true, but I'm unable to see the proof. While I could prove it when $U=\mathbb{R}^n$, but for other open sets, I do not have a proof. Although I could construct $\alpha:[0,1]\to\mathbb{R}^n$ satisfying 1, 2 and 3, I can't force it to remain in $\bar{U}$. Could you please let me know if it's true. 
Question:
Let $n\geqslant 2$, $U\subset\mathbb{R}^n$ be open, bounded, connected, $C^\infty$ with $\bar{U}$ path connected and let $x_0\in\bar{U}$. Let $(p_n)_{n\in\mathbb{N}}$ be a sequence in $\bar{U}$ such that $p_n\to p$ in $\bar{U}$. Does there exist a $C^\infty$ path $\alpha:[0,1]\to \bar{U}$ and a sequence $(t_n)$, $t$ in  $(0,1)$ such that $t_n\to t$ and


*

*$\alpha(0)=x_0$.

*$\alpha(t_n)=p_n$, and

*$\alpha(t)=p$.

 A: In my other answer, I show how the result for $\mathbb{R}^N$ can be extended to arbitrary $U$. Here I prove that the result is actually false for $U = \mathbb{R}^N$ (so in a sense my other answer is completely void).
Consider $p = 0$ and $p_n$ such that the shortest path that includes all $p_n$ contained in the shell $S_k = \{x \in \mathbb{R}^N : 2^{-k - 1} < |x| < 2^{-k}\}$ is $k$. This is easily achieved if $p_n$ is defined to be the enumeration of the set $$\bigcup_{k = 0}^\infty S_k \cap (\varepsilon_k \mathbb{Z})^N $$ for $\varepsilon_k > 0$ decreasing sufficiently fast.
If $\alpha$ is a path with the desired properties, then $\alpha$ passes through all points $p_n$ contained in $S_k$, and so the length of $\alpha$ is not less than $k$. Since $k = 0, 1, 2, \ldots$ is arbitrary, we get a contradiction: being smooth, $\alpha$ must have finite length.
A: If $p \in U$, then there is a neighbourhood $V$ of $p$ which is diffeomorphic with $\mathbb{R}^N$ and which contains all but finitely many $p_n$. Let $\Phi : \mathbb{R}^N \to V$ be such a a diffeomorphism. One can use the result for $\mathbb{R}^N$ (and $\Phi^{-1}(p_n)$, $\Phi^{-1}(p)$) to get a path $\beta : [\tfrac{1}{3}, \tfrac{2}{3}] \to \mathbb{R}^N$ which contains almost all points $\Phi^{-1}(p_n)$, and then map it back into $V$ using $\Phi$ to get $\alpha = \Phi \circ \beta$. It remains to extend $\alpha$ to $[0, 1]$ so that it includes the finitely many remaining points $p_n$ and $x_0$.

If $p \in \partial U$, then a very similar argument works: there is a (relative) neighbourhood $V$ or $p$ in $\overline{U}$ which is diffeomorphic with the half-space $\mathbb{R}^{N-1} \times [0, \infty)$. Let $\Psi$ be such a diffeomorphism, and consider $\Phi : \mathbb{R}^N \to V$ given by $$ \Phi(x_1, x_2, \ldots, x_{N-1}, x_N) = \Psi(x_1, x_2, \ldots, x_{N-1}, (x_N)^2) . $$ Now we can simply apply the argument from the previous paragraph, with an arbitrary choice of $\Phi^{-1}(p_n)$ when $p_n \in U$.
