Every $SO_3$-invariant subspace contains an $SO_2$-invariant function?

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Trying to understand a lemma involving spherical harmonics. Lemma: In every nonnull finite-dimensional $SO_3$-invariant subspace U in the space of continuous functions on the sphere there is a nonnull $SO_2$-invariant function.

The key step I don't understand is that the subspace $U_0 = \{f\in U : f(o)=0\}$ has codimension 1. See the proof below:

Proof: Fix some point $o$ (maybe the north pole), and the subgroup of $SO_3$ that fixes $o$ is isomorphic to $SO_2$. $U$ contains functions that do not vanish at $o$ because you can take any nonnull function in $U$ and find a point where it doesn't vanish, then rotate that point to $o$ via $SO_3$ and if you act on that nonnull function you can get a function in $U$ that doesn't vanish at $o$.

Define the subspace $U_0 = \{f\in U : f(o)=0\}$. It has codimension 1. It is clearly $SO_2$ invariant and so its orthogonal complement $U_0^{\perp}$ is also $SO_2$ invariant. Choose any nonnull $f_0$ in $U_0^{\perp}$, then rotations around the axis through $o$ can only affect it by a multiplicative constant, but since the point $o$ is fixed, that constant must be 1. So $f_0$ is our desired $SO_2$-invariant function.

I don't understand why the complement of $U_0$ is 1-dimensional.