Let $U$ be a subspace of $\mathbb R^n$ with dimension $d<n$. Prove that there exists a vector $v$ with non-zero coordinates such that $v\notin U$.
The fact that the vectors with zeros in one coordinate form a vector space might be useful. An idea is to assume the contrary then for each vector in $\mathbb R^n-U$ at least one coordinate is zero. Thus, infinitely many vectors can be found in $\mathbb R^n-U$ with the same zero coordinate. This can help us if we consider the $\operatorname{span}$ of these vectors and repeat the process with $\mathbb R^n-U-V_S$ where $V_S$ is the vector space of vectors consisting of vectors with zero coordinates at indices $i\in S$. So it seems sufficient to prove that we cannot Partition $\mathbb R^n$ to some subspaces $V_{S_i}$ and $U$.
Best Answer
As you noticed, the result you request is the consequence that a linear subspace over an infinite field is not a finite union of proper subspaces. This result is itself a consequence of
If a field $F$ is such that $\left|F\right|>n-1$ why is $V$ a vector space over $F$ not equal to the union of $n$ proper subspaces of $V$