Why can a closed, bounded interval be uncountable?

I suspect you're conflating two meanings of "finite". Some sets are finite, meaning they have only finitely many elements. An interval like $[0,1]$ is not such a set. On the other hand, $[0,1]$ has finite length, which is a quite different matter. As the other answers have explained, finite length does not imply finiteness (or even countability) in terms of the number of elements.

Certainly the interval $[0,1]$ is uncountably infinite; it has the same cardinality as $(0,1)$ and $\mathbb{R}$ and $\mathcal{P}(\mathbb{N})$ and many other sets which are commonly used in mathematics.

However, closed and bounded intervals such as $[0,1]$ do have a nice finiteness property called compactness: that is, if $\{ U_i \mid i \in I \}$ is any collection of open intervals such that $\bigcup_{i \in I} U_i = [0,1]$, then there is a finite subset $J \subseteq I$ such that $\bigcup_{i \in J} U_j = [0,1]$.

This is not true of $(0,1)$, since for instance if we define $U_i = (\frac{1}{i}, 1-\frac{1}{i})$ then $\bigcup_{i \in I} U_i = (0,1)$ but no finite subset $J \subseteq I$ has $\bigcup_{i \in J} U_i = (0,1)$.