Does real dimension equal rational dimension?

WLOG, we may suppose the vectors are independent in $\mathbb{Q}^n$ (since if they aren't, we can just throw away vectors until we have a basis without changing $V$ or $W$).

What you're really proving is that if the vectors are independent in $\mathbb{Q}^n$, then they are independent in $\mathbb{R}^n$. For clearly, the converse follows.

The trick is to construct an orthogonal basis. This trick works in $\mathbb{Q}^n$ because the key step of the procedure is taking a vector $v$ and replacing it by $v' = v - \frac{w(v \cdot w)}{|w|^2}$. But if $v, w \in \mathbb{Q}^n$, then the replacement $v'$ is also in $\mathbb{Q}^n$.

if you collect the $\mathbf v_j$ in a matrix, you can make this a result about polynomials.

$\mathbf V:=\bigg[\begin{array}{c|c|c|c|c} \mathbf v_1 & \mathbf v_2 &\cdots & \mathbf v_{k}\end{array}\bigg]$

working over $\mathbb Q$ (or some subfield) we have: $\text{rank}\big(\mathbf V\big) = r$

Now use the fact that a matrix has rank $r$ iff it has some $r\times r$ submatrix with nonzero determinant and for $m\gt r$ all $m\times m$ submatrices have zero determinant.

These determinants don't change when you consider $\mathbb R$, or any extension, hence the rank doesn't change either.