Suppose $U\subset \mathbb{R}^n$ is the unit ball and $\left\{x_i\right\}_{i\in \mathbb{N}}$ is a sequence in $U$ that converges to $0$. Does there exist a smooth curve $\gamma \colon \left[0,1 \right] \to U$ such that the image of $\gamma$ contains a subsequence of $\left\{x_i\right\}_{i\in \mathbb{N}}$ and such that $\gamma(0)=x_1$ and $\gamma(1)=0$?
I do not know how to approach this. The case for finitely many points already troubles me. I know about smoothing piece-wise linear curves but that is not really helpful here.
Edit: So, I did have an idea to get a $C^0$-curve that is analytic on $(0,1]$, so everywhere except one boundary point. Sadly my approach will probably never explicitly produce a $C^{\infty}$-curve.
Consider $\mathbb{R}^n\subseteq \mathbb{C}^n$ and the given sequence as a sequence in $\mathbb{C}^n$. That is one gets $n$ sequences in $\mathbb{C}$ by considering the components $\left\{ x_i^j \right\}_{i\in \mathbb{N}}$. Also one may consider the discrete sequence $\left\{1,2,3,... \right\}$ in $\mathbb{C}$. As $\mathbb{C}$ is a Stein manifold it follows that there exists holomorphic functions $f^j\colon \mathbb{C} \to \mathbb{C}$ such that
$f^j(i)=x^j_i.$
Then $g'^j:=\mathrm{Re}(f^j)\mid_{\mathbb{R}}\colon \mathbb{R} \to \mathbb{R}$ also satisfies the above equation and note that $\lim_{x\to \infty }g'^j(x)=0$.
One may moreover consider the analytic map
$g^j\colon (0,1] \to \mathbb{R},\; x \mapsto g'^j(1/x)$
it satisfies $\lim_{x\to 0} g^j(x)=0$ and thus extends continuously to $[0,1]$. Moreover, its image contains the entire sequence one started with by construction.
