Disclaimer: this is not an answer to the question, but a long comment I wish to keep separate. The proof of the result in the question follows immediately from the two lemmas below.

**Definition.** Two subspaces $L, K \subset V$ of a finite-dimensional spaces $V$ are said to have *minimal intersection* if the dimension of their intersection is the least possible compatible with their dimensions and the dimension of $V$.

**Lemma 1.** *If every subspace of $\mathbb{R}^n$ that is spanned by integer vectors has minimal intersection with the orthogonal complement of a $k$-dimensional subspace $\zeta$ $(1 \leq k \leq n-1)$, then the orthogonal projection of $\mathbb{Z}^n$ onto $\zeta$ is dense in $\zeta$.*

*Proof.* Let $D_\epsilon \subset \zeta$ be the closed disc of radius $\epsilon > 0$ centered at the origin, and let $C_\epsilon$ be the cylinder $\pi_\zeta^{-1}(D_\epsilon)$, where $\pi_\zeta : \mathbb{R}^n \longmapsto \zeta$ is the orthogonal projection onto $\zeta$. Note that $C_\epsilon$ is a $0$-symmetric, closed, convex set with infinite volume and hence contains an infinite number of
integer vectors.

Claim. The dimension of the subspace $\eta_\epsilon$ spanned by the (infinite) set of all integer vectors in $C_\epsilon$ is greater than $k$, the dimension of the subspace $\zeta$.

Indeed, $\eta_\epsilon$ is spanned by integer vectors and if its dimension were less than or equal to $k$, then by hypothesis it would intersect $\zeta^\bot$ only at the origin. This would imply that the intersection of $\eta_\epsilon$ and the cylinder $C_\epsilon$ is compact and hence it could not contain an infinite number of integer vectors. This would contradict the definition of $\eta_\epsilon$.

Notice now that if the dimension of $\eta_\epsilon$ is greater than $k$, then the minimal intersection hypothesis implies that its projection onto $\zeta$ is surjective. Hence, we can find integer vectors $\mathbf{m}_1$, $\mathbf{m}_2$, $\dots$, $\mathbf{m}_k$ in $C_\epsilon$ such that their projections form a basis of $\zeta$. The norm of these projections is less than or equal to $\epsilon$ and so the lattice spanned by these vectors is a subset of $\pi_\zeta(\mathbb{Z}^n)$ which is $2 \epsilon$ close to every point in $\zeta$. Since $\epsilon > 0$ was arbitrary, we conclude that $\pi_\zeta(\mathbb{Z}^n)$ is dense in $\zeta$.

**Lemma 2.** *The set of $k$-dimensional subspaces of $\mathbb{R}^n$ that have minimal intersection with every subspace spanned by integer vectors has full measure in the Grassmannian $G_k(\mathbb{R}^n)$*.

*Proof.* Fix a dimension $p$ between $1$ and $n-1$ and a $p$-dimensional subspace $\eta$. Consider the set of all $(n-k)$-dimensional subspaces $\zeta^\bot$ that *do not* intersect $\eta$ minimally. This is a finite union of lower-dimensional submaifolds and hence has measure zero in $G_{n-k}(\mathbb{R^n})$. The union of all these sets as $\eta$ ranges over the countable set of all subspaces spanned by integer vectors has still measure zero and so its complement has full measure.