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Here's Theorem 4.17 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Suppose $f$ is a continuous 1-1 mapping of a compact metric space $X$ onto a metric space $Y$. Then the inverse mapping $f^{-1}$ defined on $Y$ by $$ f^{-1}( f(x) ) = x \ \ \ \ \ \ \ ( x \in X ) $$ is a continuous mapping of $Y$ onto $X$.

Now in Example 4.21 Rudin illustrates that compactness of $X$ is essential in Theorem 4.17.

So here is Example 4.21:

Let $X$ be the half open interval $[0, 2 \pi )$ on the real line, and let $\mathbf{f}$ be the mapping of $X$ onto the circle $Y$ consisting of all points whose distance from the origin is $1$, given by $$ \mathbf{f}(t) = ( \cos t, \sin t ) \ \ \ \ \ \ ( 0 \leq t < 2 \pi). $$ Then this function $\mathbf{f}$ is of course continuous and injective. But Rudin asserts that the inverse function is discontinuous at the point $(1, 0) = \mathbf{f}(0)$. How to demonstrate this fact rigorously?

So far in the book, Rudin has not discussed the functions $\cos$ and $\sin$. So can we come up with an example that uses only what Rudin has discussed?

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    Basically any point slightly above $(1, 0)$ has angle $ \approx 0$, whereas those slightly below have angle $\approx 2 \pi$. From here, it's a straightforward proof of discontinuity. – 2017-01-13

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The inverse function goes from the unit circle to the interval $[0, 2\pi)$. Consider looping around the circle toward the point $(1,0)$, once clockwise and once counter-clockwise.

When you approach clockwise, your angle (the output of the inverse function) decreases to $0$. But, when you approach counter-clockwise, your angle is INCREASING to $2\pi$. So, there is not a consistent limit at that point, and it is therefore discontinuous.