finding $C^1$ path on an open and path connected set. Given an open and path connected set $U\subseteq \mathbb R^n$, is there a way to find a $C^1$ path between every $a,b\in U$? 
If so, is there a general proof of existence of such path?
 A: Given any two points in your domain, pick a continuous path between them. Because the closed interval $I$ is compact, you can cover this path with finitely many balls; use this to construct a piecewise linear path between the two points. (With enough care in this step and the previous you can ensure that the path is actually injective.) Now smooth the corners.
A: We can also use a tweaked version of the Weierstrass approximation theorem: Given $f$ continuous on $[a,b],$ there are polynomials $p_m \to f$ uniformly on $[a,b]$ such that $p_m(a) = f(a), p_m(b)=f(b)$ for all $m.$ (For the proof, see * below.)
Given this, suppose $\gamma:[0,1] \to U$ is a path from $x$ to $y$ in $U\subset \mathbb {R}^n.$ We can then choose $n$ polynomials that uniformly approximate the components of $\gamma,$ with agreement at the end points. Because the range of $\gamma $ is compact and these polynomial maps are as close to $\gamma $ desired, eventually they all map into $U$, and any one of these will be a polynomial path from $x$ to $y$ in $U.$
*Suppose first $f(a)=f(b)=0.$ Find polynomials $p_m \to f.$ (Here and below, the $\to $ sumbol means uniform convergence on $[a,b].$) Then $q_m = p_m - p_m(a) \to f$ and all $q_m(a) = 0.$ This implies $r_m(x)=q_m(x) -(x-a)q_m(b)/(b-a) \to f(x),$ and $r_m(a) = r_m(b) = 0$ for all $m.$ For the general $f,$ let $l(x)$ be the line through $(a,f(a)), (b,f(b)).$ Apply the above to find $r_m \to f-l$ with $r_m$ zero at the end points. Then $r_m+l$ does the job.
