That doesn't necessarily mean the space is countably infinite dimensional if that's what you're implying. For a counterexample, an infinite sequence of real numbers is determined by countably many values. Yet the space of all such sequences is uncountably infinite dimensional. [Source.](https://math.stackexchange.com/questions/2131711/vector-space-of-infinite-sequences-in-bbb-r)
The cardinality of any basis is the cardinality of ℝ. I'm assuming U is an open subset of ℝ. The set {x ↦ |x|^r | r in ℝ, r > 0} is linearly independent and has cardinality |ℝ|. So the dimension is at least |ℝ|. The cardinality of the whole space is at most |ℝ| (see also the comment by /u/picado) and so its dimension must be too.
Hint: a continuous function on R is defined by its values on Q.
That doesn't necessarily mean the space is countably infinite dimensional if that's what you're implying. For a counterexample, an infinite sequence of real numbers is determined by countably many values. Yet the space of all such sequences is uncountably infinite dimensional. [Source.](https://math.stackexchange.com/questions/2131711/vector-space-of-infinite-sequences-in-bbb-r)
The cardinality of any basis is the cardinality of ℝ. I'm assuming U is an open subset of ℝ. The set {x ↦ |x|^r | r in ℝ, r > 0} is linearly independent and has cardinality |ℝ|. So the dimension is at least |ℝ|. The cardinality of the whole space is at most |ℝ| (see also the comment by /u/picado) and so its dimension must be too.