Proving that the inverse of a bijective continuous function is continuous I'm trying to prove the following statement: If $f: [a,b] \rightarrow B$, where $B$ is a subset of the reals, is continuous and bijective then it's inverse $f^{-1}$ is continuous.
Firstly we note that $f^{-1}$ is monotonic since $f$ is monotonic. Say we have $f^{-1}$ is strictly increasing (the proof for it being strictly decreasing would be similar), then I have the following proof for its continuity: 

I can understand everything in the proof except from the chosen value for $\delta_{\epsilon}$. I think $\delta_{\epsilon}$ should be chosen to be $\delta_{\epsilon}=max\{y_+ - y_0 , y_0-  y_-\}$ because if we had $\delta_{\epsilon}=min\{y_+ - y_0 , y_0-  y_-\}$ and we  had that $y_+ - y_0>y_0 - y_-$ then $\delta_{\epsilon}=y_0 - y_-$ and $|y-y_0|<\delta_{\epsilon} \rightarrow y \in (y_-, y_0)$ not $y \in (y_0, y_+)$ so we also wouldn't have $y \in (y_-,y_+)$. However, I'm not sure if I'm right or not and if the proof is actually right.
 A: Your question seem to imply that $B$ is a subset of the real numbers (this statment stays true if $B$ is any Hausdorff topological space). A function $g$ is continuous if $g^{-1}(A)$ is closed for every closed set $A$. Then to show that $f^{-1}$ is continuous, we have to show that the image of every closed set is closed.
Let $A$ be a closed subset of $[a,b]$, then $A$ is compact since it is the intersection of a closed set ($A$) and a compact set ($[a,b]$). Since compactness is a topological invariant, $f(A)$ must be compact which means that $f(A)$ is closed. 
A: I believe the relevant proof in the abstract topological case is that a continuous bijection from a compact space to a Hausdorff space is a homeomorphism as can be seen here.
However regarding the proof, $B$ has to be an interval since again abstractly the continuous image of a connected set is connected, in other words intervals are mapped to intervals. It can also be shown that $B$ needs to be a closed interval, topologically you could consider compactness.
Continuous and bijective implying monotonic is using a property of the interval.
Notice that $f$ and $f^{-1}$ strictly monotonically increasing, guarantees $y_+>y_-$. Now if $\delta_\epsilon=\min\{y_+-y_0,y_0-y_-\}$ which are both positive numbers, then supposing $\delta_\epsilon=y_+-y_0$, then $|y-y_0|<\delta_\epsilon$ means $y\in (y_0-\delta_\epsilon,y_0+\delta_\epsilon)=(2y_0-y_+,y_+)\subset (y_-,y_+)$ since if we subtract the bigger number $y_0-y_-$ from $y_0$, we get $y_-$. It's similar the other way round.
I also feel like the last step is a little brief, but essentially $y\in (y_-,y_+)$ and using monotonicity guarantees 
$$f^{-1}(y_0)-\epsilon=x_0-\epsilon=f^{-1}(y_-)\leq f^{-1}(y)\leq f^{-1}(y_+)=x_0+\epsilon=f^{-1}(y_0)+\epsilon$$
