I need to prove that the $\epsilon$-$\delta$ definition of continuity implies the open set definition continuity for a real function. Here's my attempt.
For any basis $V: (a, b)$ in the range, for each $f(x) \in V$,
let $\epsilon = \min(f(x) - a, b - f(x))$, then for any $x$ that $f(x) \in V$ according the $\epsilon-\delta$ definition of continuty there must exists a $\delta$ that the open set $U_x : (x - \delta, x + \delta) \subset f^{-1}((f(x) - \epsilon, f(x) + \epsilon)) \subset f^{-1}(V)$
In conclusion, $$f^{-1}(V) = \bigcup_{x \in f^{-1}(V)} U_x .$$ $f^{-1}(V)$ is an open set. Then for any open set $W$, $$f^{-1}(W) = \bigcup_{V \subset W} f^{-1}(V)$$ $f^{-1}(W)$ is an open set. So for any open $W$, $f^{-1}(W)$ is also an open set. This is exactly the open set definition of continuty. QED.
Is my answer correct? Thanks.