I constructed a proof that may be invalid. Is this proof rigorous enough/usable?
Prove: $f$ is a function which is continuous on the interval $I$ (the range of $f$ is on the interval $J$) and $f$ has an inverse. Prove the inverse of $f$ is continuous on $J$.
Proof: If $f$ has an inverse, that means $f$ is one-to-one and also is strictly monotonic. Let's assume $f$ is always increasing. We must prove that the definition of continuity holds for all inputs on the inverse function in the interval $J$, with $y_0$ being any input on it (and $x_0$ being the corresponding $x$ value to $y_0$ on $f$):
$$\lim_{y \to y_0}f^{-1}(y) = f^{-1}(y_0)$$
We know, because $f$ is always increasing, that
$f(x_0 - \epsilon) < y < f(x_0 + \epsilon)$
for any $x_0 \in I$ and any $\epsilon \gt 0$
We also know, for some $y$ and any $\delta \gt 0$ that
$f(x_0) - \delta < y < f(x_0) + \delta$.
choose sufficiently small $\delta$ so that
$f(x_0 - \epsilon) < f(x_0) - \delta$ and $f(x_0) + \delta < f(x_0 + \epsilon)$.
So, for some $y$ such that
$f(x_0) - \delta < y < f(x_0) + \delta$.
$y_0 - \delta < y < y_0 + \delta$
$-\delta < y - y_0 < \delta$
$\vert y - y_0 \vert < \delta$,
then
$f(x_0 - \epsilon) < f(x_0) - \delta < y < f(x_0) + \delta < f(x_0 + \epsilon)$.
$f(x_0 - \epsilon) < y < f(x_0 + \epsilon)$
$x_0 - \epsilon < f^{-1}(y) < x_0 + \epsilon$
$-\epsilon < f^{-1}(y) - x_0 < \epsilon$
$-\epsilon < f^{-1}(y) - f^{-1}(y_0) < \epsilon$
$| f^{-1}(y) - f^{-1}(y_0) | < \epsilon$ for any $\vert y - y_0 \vert < \delta$.
Therefore, the limit definition of continuity holds for any arbitrary point $(y_0, x_0)$ on the inverse function, so if the function $f$ is continuous on $I$ and has an inverse, then the inverse is continuous on $J$.
Is this proof valid? Thank you in advance for any advice.