In my real analysis course, we need to prove that if $K\subseteq\mathbb{R}$ is compact, and $f:K\to\mathbb{R}$ is continous and injective, then $f^{-1}$ is continuous. The standard proof for this uses the sequential characterisation of continuous functions, and in this proof one certainly needs the domain to be closed and bounded, and one also needs injectivity, so its natural to think that we need $K$ to be compact in order for this to work. However, I seem to have a proof that doesn't require $K$ to be compact, but it uses some tools which are not taught in the course, and which I am not sure I am using correctly, I will work with the special case of $K=\mathbb{R}$ which is not compact:
To prove that $f^{-1}$ is continuous, it suffices to show that the inverse image of any open subset $A\subset\mathbb{R}$ is also open, so we need to show that
$$(f^{-1})^{-1}(A) $$
Is open. But
$$ (f^{-1})^{-1}(A) = f(A) $$
And since $f$ is continuous and injective, by the invariance of domain, $f(A)$ is open, completing the proof that $f^{-1}$ is continuous. So apparently we didn't need $K$ to be compact, but at the same time I suspect there are some choices of $K$ and $f$ that don't have this property, the proof above fails for a general subset of $\mathbb{R}$ when we try to conclude that $(f^{-1})^{-1}(A)= f(A)$, since $A$ might not be contained in the domain of $f$ this statement doesn't even make sense. What's going on here? Is there a mistake in my proof, or is there some larger class of sets for which this condition is true, and is the first proof requiring compactness misleading us?
It's been noted already in the comments by Rob Arthan that for the case $K=\mathbb R$ (or $K$ any interval for that matter) you can actually just use the intermediate value theorem to establish monotonicity, and get continuity of the inverse from there. (This is perhaps another way of saying that the proof of invariance of domain is not so hard in dimension $1$).
However, all of that breaks down when $K$ is not connected.
Without compactness or connectivity, we can let $K=[0,1)\cup [2,3)$, and then the map $f\colon K\to [0,2)$ given by $$ f(x)= \begin{cases} x & x\in [0,1)\\ x-1 & x\in [2,3) \end{cases} $$ is continuous and injective, but $f^{-1}$ is certainly not continuous.
Remark
If $K$ is any open set, your proof will work (indeed, invariance of domain essentially states the result you wish to prove), as will the IVT argument (when applied separately to the given components of $K$). Otherwise, if $A\subseteq K$ is open in the subspace topology of $K$, that merely means $A=K\cap U$ for some open $U\subseteq \mathbb R$. In this case, $f$ is not defined on all of $U$, so we cannot apply invariance of domain.
Update
Regarding the final remarks, I wouldn't say the original statement is misleading. Just because you are asked to prove that compactness implies a certain nice property, you should never assume from that that said property will always fail in every noncompact setting. If you were asked to prove that $f^{-1}$ could only be continuous when $K$ is compact, then you could justifiably feel misled.
As it is, you've happened upon two different classes of nice subsets (namely, open subsets, and connected subsets) for which the statement also works.
However, once you step outside of $\mathbb R$, the statement regarding compact sets will generalize much further than the ones for connected or open sets.