Showing that 4 implies 5 reduces to showing:
Let $A\in\mathrm{GL}_n(\mathbf{Z})$. Let $V_-,V_{\mathrm{ru}}\subset\mathbf{C}^n$ be the sum of characteristic subspaces of $A$ relative to eigenvalues of modulus $\le 1$ (resp., to eigenvalues that are roots of unity). Then $V_-\cap\mathbf{Q}^n=V_{\mathrm{ru}}\cap\mathbf{Q}^n$.
Separating roots of unity and other eigenvalues, this reduces to showing:
Let $A\in\mathrm{GL}_m(\mathbf{Z})$, assume that $A$ has no root of unity as eigenvalue. Then $V_-\cap\mathbf{Q}^m=\{0\}$.
Suppose we have a counterexample. Passing to either a smaller invariant subspace or to a quotient allows to assume that the action of $A$ on $\mathbf{Q}^m$ is irreducible. Then $V_-\cap\mathbf{Q}^m$ is clearly a invariant subspace and is nonzero. So it is all of $\mathbf{Q}^m$. Hence $A$ has all its eigenvalues of modulus $\le 1$. Since the product of its eigenvalues, given by the determinant, is $\pm 1$, all its eigenvalues have modulus 1 (and are not roots of unity). This cannot hold, by standard "geometry of numbers" argument à la Minkowski.
Here's the simple argument for the last point:
Let $A\in\mathrm{GL}_m(\mathbf{Z})$ act irreducibly on $\mathbf{Q}^m$, with only complex eigenvalues of modulus 1. Then $A$ has finite order.
Proof:The irreducibility assumption implies that the minimal polynomial of $A$ is irreducible, hence $A$ is $\mathbf{C}$-diagonalizable. Since $A$ has only eigenvalues of modulus 1, it follows that the subgroup $\langle A\rangle$ is bounded in $M_m(\mathbf{C})$. Since it lies in $M_m(\mathbf{Z})$, it is therefore finite.