Show that $f, f^{-1}$ are continuous 
Let $A,B \subset \mathbb{R}$ be open, and $f:A\rightarrow B$ be surjective and strictly monotonic increasing. Show that $f,f^{-1}$ are continuous.

Proof: I first show $f$ is injective. Let $x,y \in A, \mbox{and } x\neq y.$ This means either $x<y \mbox{  or } y<x.$ As $f$ is monotonic increasing, $f(x)<f(y) \mbox{ or }f(y)<f(x).\Rightarrow f(x)\neq f(y)\Rightarrow f$ is injective. This shows $f$ is bijective. 
To show $f$ is continuous, let $D \subset B$ be open. I need to show $f^{-1}(D)$ is open in $A$. Suppose not, i.e., $(f^{-1}(D))^{c}$ is not closed. $\quad\Rightarrow \exists$ a sequence $(x_n)_{n\in\mathbb{N}}$ in  $(f^{-1}(D))^{c}$ that converges to $x$ which is in $f^{-1}(D)$. As $f$ is bijective, $\exists$ a unique $y_n,y$ for each $x_n$ such that $f(x_n)=y_n,f(x)=y,\forall n\in\mathbb{N}$, where $(y_n)_{n\in\mathbb{N}}$ is in $D^{c}, y\in D$. Here, $y_n\rightarrow y$. If it does not, this means $f^{-1}(y_n)=x_n$ does not converge to $f^{-1}(y)=x$, contradiction.  $\quad \Rightarrow D^c$ is not closed. $\Rightarrow D$ is not open. Contradiction. 
$f^{-1}$ can also be proved to be continuous in the same way as above.
I somehow get a feeling that I am not allowed to argue $y_n \rightarrow y$ because a priori, my argument, which I think, assumes $f$ is continuous, which I have not proven yet!
Is my proof correct or my doubt? Please help me get out of this situation! 
 A: I think assuming that $y_n$ goes to y is assuming f is continuous. 
The hypothesis is that f is strictly monotonic increasing. Let $x_n$ < x be a sequence which is monotonically increasing and converging to x. Then $f(x_n) \le f(x)$ for all n. Because {$x_n$} is monotone, the sequence {f($x_n$)} is as well and must have a limit, say y $\le$ f(x).  However, y cannot be strictly  less than f(x). If it is then because B $\subset$ R is open there must be a neighborhood of y in B containing a value w such that y < w < f(x).
Consider c = $f^{-1}(w)$. Because f is monotonic we have $x_n$ < c < x for all n.  But by definition $x_n \rightarrow x$.  So there is no such c, thus no such w and lim $f(x_n) = f(x)$.
Of course the same argument can be made using any decreasing sequence  approaching x.  Finally, if there is a sequence $a_n \rightarrow$ x which is not monotonic, you can repeat this argument with the lim sup and lim inf of $a_n$. So the sequence f($a_n$) must have a limit, and it must be f(x).
The conclusion is that f is indeed continuous.  
If f is continuous so is $f^{-1}$, which follows from the definition of the inverse.  Since Pete Clark is concerned about this statement, here is another approach.
Because f is monotone, $f^{-1}$ must be also.  By definition $f^{-1}(B) = A$ and we are given that A is open.  So the argument above can be applied to $f^{-1}$, making it continuous.
A: A conceptual proof follows from the material in $\S$ 6.3 of my honors calculus notes:
Step 1: We are given that $f$ is bijective and increasing.  So $f^{-1}$ exists and is moreover increasing: suppose not; then there are $y_1 < y_2 \in B$ with $f^{-1}(y_1) \geq f^{-1}(y_2)$.  Then applying $f$ we get $y_1 \geq y_2$, contradiction.  Thus the situation is perfectly symmetrical with respect to $f$ and $f^{-1}$, so it suffices to show that $f$ is continuous.
Step 2: We use the fact that every monotone function defined on $A$ has a left hand limit at every $c \in A$ -- namely $\sup \{ f(x) \mid x < c\}$ and a right hand limit -- namely $\inf \{ f(x) \mid x > c\}$, and the value $f(x)$ lies in between. (This is part of the Monotone Jump Theorem in $\S$ 6.3 of my notes.)  Thus the only way we can have a discontinuity is if $\lim_{x \rightarrow c^-} f(x) < f(c)$ or $f(c) < \lim_{x \rightarrow c^+} f(c)$.  But if either of these occurs, then $f(c)$ is not an interior point of $f(A)$, contradicting the hypothesis that $f(A)$ is open.
This answers the OP's question.  I claim that it also proves that the inverse of a continuous function is continuous, at least in the case that the domain of $f$ is an interval.  This is because every injective continuous function $f: I \rightarrow \mathbb{R}$ must be monotone: see $\S$ 5.6.3 of the notes.  (There is a bit of combinatorial trickiness here.)  Using the fact that every open subset of $\mathbb{R}$ is a disjoint union of open intervals -- which is not in the notes (I don't do any explicit topology whatsoever there) but is well known and not hard to show -- and that for every open interval $I$ and continuous $f$, $f(I)$ is an interval ($\S$ 6.2 of the notes) one sees that this extends to continuous functions on any open subset of $\mathbb{R}$, but this seems to be the longer way around this particular question.
Added: It is certainly not the case that any continuous bijection $f: X \rightarrow Y$ of topological spaces must have a continuous inverse.  To get a counterexample, let $Y$ be your favorite non-discrete topological space, let $X$ be the same set endowed with the discrete topology, and let $f$ be the identity map.  From this perspective the "automatic continuity" property of the inverse for continuous bijections on open subsets of $\mathbb{R}$ is surprising.  It can be generalized to open subsets of $\mathbb{R}^n$ and then becomes a quite famous (and rather deep) theorem, Brouwer's Invariance of Domain.  This can be generalized to topological manifolds.  There are other "Open Mapping Theorems" in mathematics -- famously in complex analysis and Banach space theory -- but such results are highly prized, as they are the exception rather than the rule.
