It's even more direct. Fix $g\in\mathrm{GL}_d(\mathbf{Z})$, acting on $T=\mathbf{R}^d/\mathbf{Z}^d$, with no eigenvalue that is a root of unity.

Let $T$ be the given torus, $T_{\mathrm{cont}}$ the subgroup of elements contracted by $g$. Define $S$ as the quotient torus $T/\overline{T_{\mathrm{cont}}}$. Then $g$ acts on the torus $S$ with no eigenvalue (on the tangent space) of modulus $\le 1$ (since by definition it has no eigenvalue of modulus $<1$, no root of unity, and algebraic units of modulus 1 have modulus $\neq 1$). Since the determinant of $g$ on (the tangent space of) $S$ is $\pm 1$, we deduce that $\dim(S)=0$, i.e. $\overline{T_{\mathrm{cont}}}$ is dense in $T$.