We call a set $S\subset E$ "absorbing" if for every $y\in E$, there exists a $t>0$ s.t. $ty\in S$. The notation $[x,y]\doteq\{z\in E: z=\alpha x+(1+\alpha y),\text{for some }\alpha\in[0,1]\}$.
I cannot understand the equivalence. If $x\in\text{core}(S)$, then $S-x$ is absorbing. This implies for every $y\in E$, there exists a $\epsilon>0$ s.t. $\epsilon y\in S-x$, i.e., $x+\epsilon y\in S$. But how can we conclude $[x-\epsilon y,x+\epsilon y]\subset S$?
Any point on the line segment is of the form $s(x-\epsilon y)+(1-s)(x+\epsilon y)$ with $0 \leq s \leq 1$. This is same as $x+(1-2s)\epsilon y$. If $s=\frac 1 2$ there is nothing to prove. If $s<\frac 12 $ use the fact that $ty \in S$ for sime $t>0$ and choose $\epsilon$ to be $\frac t {1-2s}$. If $s >\frac 1 2$ note that $t'(-y) \in S$ for some $t'>0$ and choose $\epsilon$ to be $\frac {t'} {2s-1}$.
[The converse part is almost immediate from the definitions].