Is the column vector of a uniformly sampled random unitary matrix a uniformly sampled random state vector?

Yes.

A uniformly (Haar random) sampled state vector $|psirangle$ is characterized by the fact that the probability measure is invariant under any $U$, i.e., colloquially, $U|psirangle$ is just as likely as $|psirangle$ for any unitary $U$.

On the other hand, a Haar random unitary $V$ is defined the same way: "$UV$ is just as likely as $V$, for any $U$."

Thus, a column of $V$, let's call is $|chirangle$, has again the property that $|chirangle$ and $U|chirangle$ for any $U$ are "equally likely", that is, $|chirangle$ is distributed Haar random.

(As stated in the beginning, the colloquial "equally likely" should be understood as an invariance property of the measure.)

Suppose that was not the case. Then taking the first column of a uniformly random unitary matrix gives you a nonuniformly random state.

That means that there is some state, call it $|psirangle$, that is relatively more likely to be found when sampling states with such procedure. But that would mean that the unitaries whose first column is $|psirangle$ are relatively more likely to be found than others. This is clearly against the assumption that the unitaries are uniformly sampled. We conclude that taking the first (or any other) column of a uniformly sampled unitary matrix gives a uniformly sampled quantum state.