For any function, given an interval with closed/open brackets, does a function have to have a min and max for the Extreme Value Theorem to apply? For example, does the extreme value theorem apply for the function f(x) = |x|?
In my opinion, I believe not because this function doesn't have a discrete maximum value. Is this thinking correct?
In addition, if at some point of any function is discontinuous within a interval, does it automatically mean that the application of the EVT is invalid? For example, take a simple floor function, since it's not continous, then it doesn't apply.
The EVT applies exactly when the hypothesis I mentioned in my comment exist, and there is no redundancy, meaning all of them must apply in order for the theorem to work. Importantly, it has nothing to do with differentiability.
$f(x)=\lvert x\rvert$ is indeed continuous so, pick a bounded, closed interval (say, $[a,b]$) then indeed, the EVT applies. Namely, the extreme values are $0$ if $ab<0$ or $min(\lvert a\rvert,\lvert b\rvert)$ else, for the minimum; $max(\lvert a\rvert,\lvert b\rvert)$ for the maximum.
Again, for a discontinuous function such as the floor function the EVT does not apply, which doesn't mean that, by chance, the extreme values are not attained, which indeed they are, for any bounded interval.