Are there path-connected but not polygonal-connected sets? This question came up in my mind. 

In the scope of normed spaces,  does there exist a path-connected but not polygonal-connected set? 

I'd rather say no for open sets (my intuition is that they're 'locally convex'). 
Thanks for your suggestions. 
 A: The statement is not true for arbitrary path-connected sets.  For example, the unit circle in $\mathbb{R}^2$ is path-connected, but no two points are polygonal-connected.
However, your intuition is correct for open, path-connected sets.  To see this, let $U$ be open and path connected, and let $a,b\in U$ be any two points, with $\gamma:[0,1]\to U$ injective and continuous such that $\gamma(0)=a$ and $\gamma(1)=b$.  Let $p_0=a$, $t_0=0$, and let $d_0=\mathrm{dist}(p_0,U^c)$.  Then define $B_0$ to be the ball centered at $p_0$ with radius $d_0$.  Then, inductively define $p_n$ to be the point $\gamma(t_n)$ where $t_n$ satisfies $1>t_n>t_{n-1}$ and is the smallest value $t$ in this region such that $\gamma(t)$ is on the boundary of $B_{n-1}$.  If no such $t_n$ exists, then let $p_n=b$.  Similarly, define $d_n=\mathrm{dist}(p_n,U^c)$ and $B_n$ to be the ball centered at $p_n$ of radius $d_n$.
If this process ever stops with $p_n=b$, then by the convexity of open balls, the polygonal path $p_0,p_1,\ldots,p_n$ is a polygonal path connecting $a$ and $b$ contained in $U$.  So we must show that the process stops.
If not, then there are infinitely many $p_i$ which must have a limit point $p$ in $\mathrm{im}(\gamma)$.  Let $t\in[0,1]$ be such that $\gamma(t)=p$.  Let $B$ be centered at $p$ with radius $d=\mathrm{dist}(p,U^c)$.  Notice by construction, we must have $t_n<t$ for all $n$.  Since the $p_i$ approach $p$, we may find a $p_m$ such that $\mathrm{dist}(p_m,p)<d/2$.  Since $p\notin B_m$ (else $t_{m+1}>t$), we have $d_m<d/2$.  But we also have $d_m=\mathrm{dist}(p_m,U^c)>d/2$ because $B\subset U$.  This is a contradiction.
