I have proved the following statement and I would like to know if my proof is correct and/or/if/how it can be improved.
"Suppose $f:\mathbb{R}\to\mathbb{R}$ is a strictly increasing function.
Prove that the inverse function $f^{-1}:f(\mathbb{R})\to\mathbb{R}$ is a continuous function."
My proof:
Let $f:\mathbb{R}\to\mathbb{R}$ be a strictly increasing function: then it is injective and as a function $f:\mathbb{R}\to f(\mathbb{R})$ it must be surjective so it has an inverse $f^{-1}:f(\mathbb{R})\to\mathbb{R}$ which must be strictly increasing too$^{(1)}$.
Suppose now that $f^{-1}$ were discontinuous at a point $y_d\in f(\mathbb{R})$: then, being an increasing function, $y_d$ must be a jump discontinuity so the interval $I_{y_d}:=(\lim\limits_{y \to y_d^-,\ y\in f(\mathbb{R})\\}f^{-1}(y),\lim\limits_{y \to y_d^+,\ y\in f(\mathbb{R})}f^{-1}(y))=(\sup_{y<y_d,\ y\in f(\mathbb{R})}f^{-1}(y),\inf_{y>y_d,\ y\in f(\mathbb{R})} f^{-1}(y))$ must be nonempty and we can pick an element $\bar{x}\neq x_d=f^{-1}(y_d)$ in it so $f(\bar{x})=y_d$, but being $f$ strictly increasing by hypothesis it is also either $f(\bar{x})>y_d$ or $f(\bar{x})<y_d$, a contradiction.
So, $f^{-1}(\mathbb{R})\to\mathbb{R}$ cannot be discontinuous at any point i.e. it must be continuous on $f(\mathbb{R})$. $\square$
$^{(1)}$ let $y_1,y_2\in f(\mathbb{R})$ and suppose wlog $y_1<y_2$: then $f^{-1}(y_1)=f^{-1}(f(x_1))=x_1$ and $f^{-1}(y_2)=f^{-1}(f(x_2))=x_2$ and if $x_1\geq x_2$ then $f(x_1)=y_1\geq y_2=f(x_2)$ contradiction, so it must be $x_1=f^{-1}(y_1)<x_2=f^{-1}(y_2)$
Best Answer
Consider the function $g:\Bbb R\to\Bbb R$ given by $$g(y):=\begin{cases}y-1 & y<0\\y & y\ge 0.\end{cases}$$
Clearly, $g$ is strictly increasing, is a bijection from the range of its inverse to $\Bbb R,$ and has a jump disontinuity at $y=0.$ Letting $y_d=0,$ we see that $$\left(\lim_{y\to y_d^-,\ y\in\operatorname{dom}g}g(y),\lim_{y\to y_d^+,\ y\in\operatorname{dom}g}g(y)\right)=(-1,0)$$ is certainly nonempty, but contains no elements of the range of $g$--that is, no element of the domain of the inverse of $g.$
That's the flaw in your argument. Just because an interval is non-empty doesn't mean that it contains an element in the domain of an arbitrary, strictly increasing function. Since all you've concluded about $f^{-1}$ is that it is strictly increasing, is a bijection from the range of its inverse to $\Bbb R,$ and it has a jump discontinuity, then it is entirely possible that $f^{-1}=g,$ in which case your argument falls down.
In other words, you haven't actually justified the statement
so haven't obtained your contradiction.
Added: A better approach would be to proceed directly. Take an arbitrary $y_0\in f[\Bbb R],$ and let $x_0:=f^{-1}(y_0).$ Since $f$ is strictly increasing, then for $x<x_0$ (resp., for $x>x_0$) we have $f(x)<y_0$ (resp. $f(x)>y_0$).
Take an arbitrary $\varepsilon>0,$ let $y_m:=f(x_0-\varepsilon),$ and let $y_M:=f(x_0+\varepsilon),$ so that $y_m,y_M\in f[\Bbb R]$ and $y_m<y_0<y_M.$
Letting $\delta=\min\{y_0-y_m,y_M-y_0\},$ we have $\delta>0,$ and for all $y\in\Bbb R,$ if $|y-y_0|<\delta,$ then $y_m<y<y_M.$
In particular, take any $y\in f[\Bbb R]$ such that $|y-y_0|<\delta,$ and let $x=f^{-1}(y).$ Since $f$ is strictly increasing and $f(x_0-\varepsilon)=y_m<y=f(x),$ then $x_0-\varepsilon<x.$ Similarly, $x<x_0+\varepsilon,$ and so $|x-x_0|<\varepsilon,$ or equivalently, $$\bigl|f^{-1}(y)-f^{-1}(y_0)\bigr|<\varepsilon,$$ whence we have showed that $f^{-1}$ is continuous at $y_0,$ as desired.