= Solution
Now $T^2=S=\operatorname{diag}(1,i)$. The gates $H,S$ and CNOT are <Clifford gates>: conjugating a <Pauli matrix> or a <tensor product> of them gives another signed Pauli tensor. On one wire,
$$
HXH=Z,\quad HZH=X,\quad SXS^\dagger=Y,\quad SZS^\dagger=Z.
$$
For CNOT the generator images are $X\otimes I\mapsto X\otimes X$, $I\otimes X\mapsto I\otimes X$, $Z\otimes I\mapsto Z\otimes I$, and $I\otimes Z\mapsto Z\otimes Z$. These identities can be checked on the four computational <vectors> and generate the preservation property for all Pauli tensors.<The Heisenberg Representation of Quantum Computers by Daniel Gottesman>[Section 3 and Table 1, Clifford generators and their Pauli conjugation rules]{ref}
For a fixed number $n$ of wires there are only finitely many possible images of the $2n$ generators $X_j,Z_j$ in the <Pauli group>. Their images determine a unitary up to <global phase>. To see this, if two <unitary operators> induce identical images, their quotient commutes with every $Z_j$, so is diagonal in the <computational basis>; commuting with every $X_j$ then forces all diagonal entries to be equal. The quotient is a scalar phase. Thus the generated gates form a finite set modulo <global phase>, which cannot be dense in the continuous space of all <unitary operators>. This is <finiteness of the projective Clifford group>.
A small visible consequence is that from $|0\rangle$, one-qubit <Clifford circuits> reach only the six states $|0\rangle,|1\rangle,|+\rangle,|-\rangle,|+i\rangle,|-i\rangle$ up to <global phase>. For instance, $T|+\rangle$ is outside this finite set. The fixed-$n$ Pauli argument also covers circuits using CNOT on several wires. Therefore \b[$\{\mathrm{CNOT},H,T^2\}$ is not universal].
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