If $f:[a,b] \rightarrow [c,d]$ is a continuous bijective function, then prove $f^{-1} $ (its inverse function) is continuous on $[c,d]$.
I know this can be proven by using monotony of $f$, can anyone help me finish my approach? $\textbf{My approach: }$ Let $s_n \rightarrow s $ on $[c,d]$. And let $ t_n = f^{-1}(s_n),$ and $t=f^{-1}(s)$. We have to show that $t_n \rightarrow t$.
By Bolzano-Weierstrass Theorem, there exists a subsequence $t_{n_k} \rightarrow r $ for some $r \in [c,d]$. So we have $$t_{n_k} \rightarrow r$$ $$f(t_{n_k}) \rightarrow f(r) $$ $$s_{n_k} \rightarrow f(r)$$ But since $s_{n_k}$ has $s$ for its limit, then we have $ f(r) = s \implies r=f^{-1}(s) = t $. So we know that $$t_{n_k} \rightarrow t$$ How can I conclude that in fact, every term of $t_n$ goes to t? I tried asuming the existence of another sequence $s_{n_p}$ that converges to $s$, but $f^{-1}(s_{n_p})$ does not go to $f^{-1} (s)$ and reaching a contradiction. Or alternativley one might try proving $t_n$ is a Cauchy sequence. Thanks.
Assume $t_n$ does not go to $t$. Since $t_{n_k}$ does, then $t_n$ has a another subsequence $t_{n_j}$ that is at least $\delta>0$ away from $t$ for every $n_j$. But $t_{n_j}$ is in a compact space, so (by Bolz.W.) $t_{n_j}$ has $t_{n_{j_m}}$ that goes to $p$, which cannot be $t$. Since $f$ is continuous, then $f(t_{n_{j_m}})$ goes to $f(p)$, and since $f(t_{n_{j_m}})$ is in $f(t_n)$, then $f(p)=f(t)$. But $f$ is injective, so $t$ must be $p$. Contradiction.