Does this seem like a reasonable definition of a uniform distribution?
Let \begin{align} &S \text{ be a sample space}; \tag{1} \\ &X:S\rightarrow \mathbb{R} \; \text{ be a random variable on } S. \tag{2} \\ \end{align} Then \begin{align} & X(S) \text{ is distributed uniformly } \\ \Leftrightarrow &\text{ every realisation of } X \text{ has pre-image of equal size}, \tag{3.1}\\ \Leftrightarrow &\forall \; x_1,x_2 \in X(S): \; |X^{-1}(\{x_1\})|=|X^{-1}(\{x_2\})|. \tag{3.2} \end{align}
This is far from true in general, but it is true in the very special case where $S$ is finite and the probability measure $P$ on $S$ is uniform. Note that a probability space $S$ is not just a set, but is $S$ together with a probability measure $P$ on $S$ (meaning on a fixed $\sigma$-algebra on $S$.) The distribution of $X$ depends on the probability measure you put on $S$.
Assume that a) $S$ is a discrete probability space in the sense that $P(x) > 0$ for all $x \in S$ and b) that $S$ itself is uniform, so that $P(x) = P(y)$ for all $x,y \in S$. (We write $P(x)$ to mean $P(\{x\})$.)
If a) holds, then finding probabilities reduces to taking sums instead of needing to integrate more generally. If a) and b) both hold then $S$ is finite, since otherwise the probabilities of elements in $S$ will sum to $\infty$, not $1$. Say $|S| = n$, so that $P(x) = 1/n$ for any $x \in S$. In this special case, computing probabilities reduces to computing finite sums. We have \begin{align*} P(X = x) &= \sum_{\omega: X(\omega) = x} P(\omega) \\ &= \sum_{\omega: X(\omega) = x} 1/n \\ &= \frac{1}{n} |\{\omega: X(\omega) = x\}\end{align*}
which is the same for each $x$ since $\{x\}$ has cardinality $1$. It follows that $X$ is uniformly distributed.
If either a) or b) fail, this is not true. If $S$ has any elements of probability $0$, as in a continuous space, then cardinality doesn't have much to do with probability. In that case, you need to integrate with respect to whatever probability measure is put on $S$ instead of only talking about sums. If b) fails, then of course the cardinality of the inverse image doesn't tell you anything about the probability, since some inverse images are more likely than others.