= Solution
Squaring the diagonal gate gives
$$
V^2=\operatorname{diag}(e^{-i\pi/8},e^{i\pi/8})=e^{-i\pi/8}T.
$$
Thus the <T gate> is available up to a <global phase>. Reading the supplied W columns in order $00,01,10,11$, it sends $|01\rangle$ to $|11\rangle$, $|11\rangle$ to $|01\rangle$, and fixes the other two <vectors>: it is CNOT with the \i[second] wire controlling the first. Conjugating both wires by the <Hadamard gate> reverses that orientation:
$$
\boxed{\mathrm{CNOT}_{1\to2}=(H\otimes H)W(H\otimes H).}
$$
For instance, write $W=I\otimes|0\rangle\langle0|+X\otimes|1\rangle\langle1|$. Conjugation uses $HXH=Z$ and $H|0\rangle=|+\rangle$, $H|1\rangle=|-\rangle$, yielding $I\otimes|+\rangle\langle+|+Z\otimes|-\rangle\langle-|$. Expanding the first-wire Z eigenspaces gives $|0\rangle\langle0|\otimes I+|1\rangle\langle1|\otimes X$, as required.
We have H directly and finite constructions of CNOT and T up to physically irrelevant <global phase>. By <universality transfer by exact gate synthesis>, the assumed universality of $\{CNOT,H,T\}$ proves that \b[$\{V,H,W\}$ is universal under the phase-insensitive convention]. There is no assertion of arbitrary strict operator-norm phases: on a fixed register these generators have only a <finite set> of possible determinants, which would obstruct density in the entire unitary group if <global phase> were treated as an observable target.
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