For a fucnton defined as $f:(0,1)\rightarrow \mathbb{R}$ (Continuous and strictly increasing), we also have $f^{'}:[0,1]\rightarrow \mathbb{R}\cup(-\infty,\infty).$ Why is the set $f^{-1}((-\infty,d))=[0,f^{-1}(d))$ open in $[0,1]?$ 

Because from how I understand open sets, for some open set $U$ in $X,$ how can $U=[0,f^{-1}(d))$ be open in $[0,1]?$ For instance if $0\in U$, then we cant always find some open ball $B(0,\epsilon)\subset U,$ cause then it will lie outside of $0,$ and that portion is no longer in $U$, so what am I missing here?