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Using the result of Zajíček or Anderson-Klee https://mathoverflow.net/a/354985/121665 one can actually prove a much stronger result:

Theorem. The set $S^{n-1}\setminus U$ can be covered by countably many Lipschitz images of $\mathbb{R}^{n-2}$. In particular the het has Hausdorff dimension $n-2$.

This is consistent with the observation made by the OP that when $n=2$, dimensional case the set is countable.

This is a consequence of the following result from

Ewald, G.; Larman, D. G.; Rogers, C. A., The directions to the line segments and of the r-dimensional balls on the boundary of a convex body in Euclidean space, Mathematika, Lond. 17, 1-20 (1970). ZBL0199.57002.

The above statement is copied from the monograph.

Schneider, Rolf, Convex bodies: the Brunn-Minkowski theory, Encyclopedia of Mathematics and its Applications 151. Cambridge: Cambridge University Press (ISBN 978-1-107-60101-7/hbk; 978-1-139-00385-8/ebook). xxii, 736 p. (2014). ZBL1287.52001.

Using the result of Zajíček https://mathoverflow.net/a/354985/121665 one can actually prove a much stronger result:

Theorem. The set $S^{n-1}\setminus U$ can be covered by countably many Lipschitz images of $\mathbb{R}^{n-2}$. In particular the het has Hausdorff dimension $n-2$.

This is consistent with the observation made by the OP that when $n=2$, dimensional case the set is countable.

Using the result of Zajíček or Anderson-Klee https://mathoverflow.net/a/354985/121665 one can actually prove a much stronger result:

Theorem. The set $S^{n-1}\setminus U$ can be covered by countably many Lipschitz images of $\mathbb{R}^{n-2}$. In particular the het has Hausdorff dimension $n-2$.

This is consistent with the observation made by the OP that when $n=2$, dimensional case the set is countable.

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Using the result of Zajíček https://mathoverflow.net/a/354985/121665 one can actually prove a much stronger result:

Theorem. The set $S^{n-1}\setminus U$ can be covered by countably many Lipschitz images of $\mathbb{R}^{n-2}$. In particular the het has Hausdorff dimension $n-2$.

This is consistent with the observation made by the OP that when $n=2$, dimensional case the set is countable.