Is my following proof of the continuity of a monotonic increasing bijective function correct? Let $f: A \to B$ be a function where $A$ and $B$ are intervals in $\mathbb{R}$. I wanted to prove that if $f$ is bijective and (strictly) monotonic increasing, $f$ is also continuous for every $x_0 \in A$ (I know that every bijective function that is monotonic is also strictly monotonic, but I didn't need that for the problem I had to solve).
In our lecture, we learned that $f$ is called continuous for $x_0 \in A$ iff for every neighboorhood $V$ of $f(x_0)$ there exists a neighboorhood $U$ of $x_0$ such that $f(U) \subseteq V$. According to the definition of a neighboorhood, there is an open ball $V_B \subseteq V$ with $f(x_0)$ as its center. We define $b := \inf V_B$ and $t := \sup V_B$. Let $U := (f^{-1}(b), f^{-1}(t))$ ($f^{-1}$ exists because $f$ is bijective per definition), then we get:
$a \in U \Longleftrightarrow f^{-1}(b) < a < f^{-1}(t) \Longleftrightarrow b < f(a) < t \Longleftrightarrow f(a) \in V_B$.
Hence, $U = V_B$. This implies that $U \subseteq V_B \subseteq V$ and we thus know that $U \subseteq V$.
Did I make any mistakes or was this proof correct?
Thanks in advance for your feedback!
 A: My following proof is not correct in the case that B is an open interval, since in that case if $ B = (c, d)$ and $a \in (c, d) $ then $[a, d)$ is closed in the subspace topology and $f^{-1} (d) \not \in A$. I'll leave the post here while I think about it.
Let $[a, b]$ be any closed interval in $B$. 
Then $f^{-1} (a)$, and  $f^{-1} (b) \in A$. 
If $x \in [f^{-1} (a), f^{-1} (b)]$ then by monotonicity $f(x) \in [a, b] \implies f([f^{-1} (a), f^{-1} (b)]) \subset [a, b]$. Since $f$ is a bijection then this gives $[f^{-1} (a), f^{-1} (b)] \subset f^{-1} [a, b]$.
Conversely if $y \in [a, b] $ then again by monotonicity $f^{-1}(y) \in [(f^{-1} (a), f^{-1} (b)] \implies f^{-1} [a, b] \subset [f^{-1} (a), f^{-1} (b)] $
So, $f^{-1} [a, b] = [f^{-1} (a), f^{-1} (b)] $
I.e. the pre-image of a closed interval in B is a closed interval in A so $f$ is continuous. (see http://www.mathcs.org/analysis/reals/cont/topcont.html)
Note: for any function, if $C \subset D$ then $f(C) \subset f(D)$ The bijection condition is needed to say that $f^{-1}(f(C)) = C$, i.e. in this case that $f^{-1}(f[(f^{-1} (a), f^{-1} (b)])) = [f^{-1} (a), f^{-1} (b)]$
A: The crucial hypothesis is bijective. If f has a jump discontinuity at some point a then  the range of f would miss all values between the left hand and the right hand limits at a  which contradicts the fact that the range of  f is an entire interval. Drawing a picture might help.
