$f$ is continuous $ \iff $ $f^{-1}$ is continuous?

Is the following true?

Let $A, B \subseteq \Bbb R$ and let $f : A \to B$ be a bijective map. Then

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;f$ is continuous on $A$ $ \iff $ $f^{-1}$ is continuous on $B$

It seems like such an obvious result but I can't seem to prove it. At least is the result true if we restrict $A$ and $B$ to be intervals?

My Attempt at a Proof:

$\implies $: Let $b \in B$. I was going nowhere after considering an arbitrary sequence $ (x_n) $ in $B$ which converges to $b$. So I instead assumed that $V$ was any neighbourhood around $f^{-1}(b)$. Now suppose there is no neighbourhood $U$ of $b$ such that $ x \in U \implies f^{-1}(x) \in V $. But for every neighbourhood $U'$ of $b$ there is a neighbourhood $V'$ of $f^{-1}(b)$ such that $ f^{-1}(x) \in V' \implies x \in U' $. This is by considering $b = f(f^{-1}(b))$ and since $f$ is continuous. If one of these $V'$s is a subset of $V$ then we are done. But suppose not. I cannot proceed further. I considered arbitrarily small neighbourhoods $U'_n = \{ x \ | \ |x - b| \lt \frac 1 n \}$ but still got nowhere.

Any help is appreciated.

It is not true in general. For example, consider $A = [0, 1) \cup [2, 3]$ and $f(x) = x$ on $[0, 1)$ and $f(x) = x - 1$ on $[2, 3]$. Clearly, $f$ is continuous and bijective onto $[0, 2]$. The inverse function $g\colon [0, 2] \to A$ is defined by $g(x) = x$ for $x \in [0, 1)$ and $g(x) = x + 1$ for $x \in [1, 2]$. Clearly, $g$ is not continuous.

If you require that $A$ and $B$ are intervals, then the assertion is correct. To prove this, let $f\colon A \to B$ be continuous and bijective (where $A$ and $B$ are intervals). Then $f$ is strictly monotonous, which can easily be shown using the intermediate value theorem. But the inverse of a strictly monotonous function on an interval is continuous, which proves the assertion.

I can elaborate more on that, if you want.

In general it is not true, but you can show that (supposing $f$ continuos) for every point in which $\lim_{x\to x_0} f^{-1} (x)$ exists, then it equals $f^{-1}(x_0)$.

In fact,

Let $f$ be continuos.

We want to prove $\lim_{x \to x_0} f^{-1}(x) = f^{-1}(x_0)$.

Now, $\lim_{x \to x_0} f^{-1}(x) = (f^{-1} \circ f)(\lim_{x \to x_0} f^{-1}(x))$

But given that $f$ is continuos, we have $f^{-1}( \lim_{x \to x_0} f(f^{-1}(x)) = f^{-1}( \lim_{x \to x_0} x) = f^{-1}(x_0) $ and we are done.

Not necessarily. For instance, take the surjective continuous map from the Cantor set to the segment $[0,1]$. It is not bijective but the restriction to a subset $C'$ is. However, the inverse $[0,1]\to C'$ is not continuous since $C'$ is not connected.

It's actually false, which is why you had trouble proving it. Let $A = \mathbb{N}$, and let $B = \{0\} \cup \{ \frac{1}{n} : n \in \mathbb{N}^*\}$. Finally let $f : A \to B$, $f(0) = 0$ and $f(n) = \frac{1}{n}$. Then $f$ is bijective and continuous, but it's inverse is not continuous ($B$ has an accumulation point, $A$ doesn't).