Why is the probability of a set of outcomes possibly non-zero if each outcome's probability is zero in continuous probability? In continuous probability, $\text{Pr}(X=x) = 0$ for all $x$. However, $\text{Pr}(a \leqslant X \leqslant b) >= 0$ given $b > a$. The $>$ is key here: each individual outcome has zero probability but the probability that an outcome occurs in a SET of outcomes can be non-zero.
That the probability of each element of a set is zero but the cumulative probability of the set can exceed zero is quite paradoxical.  What is this called and how is it resolved? Is it simply a property of probability limits by assumption?
(Posting this here rather than MathExchange as it is a statistical theory question)
 A: This is a consequence of the properties of integration, which is a large sum of infinitesimally small areas.  When dealing with a continuous random variable $X$, the probability that it falls within an interval is given by the integral of the probability density over that interval:
$$\mathbb{P}(a \leqslant X \leqslant b)
= \int \limits_a^b f_X(x) \ dx.$$
One of the properties of the integral is that if you integrate over a single point, you get an infinitesimal value, which is indistinguishable from zero within the normal number system.  To understand this property, you would need to learn a bit about integrals and infinitesimals, which is something covered in calculus courses (though coverage of explicit use of infinitesimal methods is something usually reserved for specialist courses).  One interpretation for the phenomena you are examining is that an integral is a large sum of infinitesimally small areas with base length $dx$ and height $f_X(x)$ at each point $x$.  Taking $dx$ to be an infinitesimally small quantity, we can then think about the integral explicitly as a sum of infinitesimally small quantities:
$$\mathbb{P}(a \leqslant X \leqslant b)
= \sum_a^b f_X(x) \cdot dx.$$
One of the properties of the integral is that if you integrate over a single point, you get an infinitesimally small value, which is indistinguishable from zero within the standard number system:
$$\mathbb{P}(a \leqslant X \leqslant a)
= f_X(a) \cdot dx = 0.$$
(The second equals sign here reflects the transition from considering infinitesimal quantities to transferring back to the domain of the standard number system.)  The resolution of the issue you are considering comes from recognising that large sums of infinitesimals can be non-infinitesimal.  When looking at these from the perspective of the standard number system (where infinitesimals are indistinguishable from zero) this means that an integral can be zero over a point but non-zero over a larger interval.

Note for mathematics pendants: In the above exposition I am attempting to give an intuitive answer for a non-specialist, so I am glossing over the transition between standard calculus treatment within the real number system (e.g., using the Reimann integral) and the treatment of same within a hyperreal number system that includes explicit infinitesimals.  See Keisler (2022) for technical details building up the integral with explicit infinitesimals (esp. pp. 59-64).
A: 
What is this called and how is it resolved?

It's called "lack of uncountable additivity". The property that for disjoint sets $A_i$
$$
P\left[\bigcup_{i\in S} A_i\right] = \sum_{i\in S} P[A_i]
$$
(additivity) is required among the axioms of probability only when $S$ is a finite or countable family of sets.
Why that? Because we only need countable additivity to prove all results that you know in probability. Requiring uncountable additivity would give you a more limited theory, since you cannot include in your theory continuous probability measures: it would not be possible to treat models such as the Lebesgue measure on $[0,1]$ and Gaussian distributions. These are arguably only abstract models, but they have a role in approximating real-world experiments.
So it's not a bug, it's a feature that allows for a richer probability theory and more setups that satisfy the axioms of probability.
A: I can't really say much about the mathematics of infinitesimals and the probabilities of points in a continuous space, but a trivial answer is that even though the probability of a point in that continuous space is defined as being zero, as long as there is an integral of your set of points (i.e. the points are contiguous) that integral will be non zero.
Yes, that answer is probably not helpful, but this next bit might be. The question is only relevant in theory because in practice you cannot observe or record or compare any result with infinite precision. In practice the real-world analogue of a continuous distribution is unavoidably granular and the probability of each point in that granular distribution will be non-zero.
