Constructing a vector space over $\mathbb{F}$ of dimension $r(n-r)$ - how? 
Let $n ∈ N$ and $1 ≤ r ≤ n$. Denote the collection of all $r$ dimensional subspaces of $\mathbb{F}^n$ by $\mathcal{G}_r$, i.e. $\mathcal{G}_r = \{A: A\leq \mathbb{F}^n,dim(A)=r\}$. Show that, for every $W ∈ \mathcal{G}_r$, there exists $U ⊆ \mathcal{G}_r$ containing $W$ such that $U$ can be made a vector space over $\mathbb{F}$ of dimension $r(n − r)$.

Notation: $P \leq Q$ means $P$ is a subspace of $Q$.
I really don't know where to start, and I'd appreciate any help. In particular, I don't understand how to find $U$ for a given $W$ such that $U$ can be made into the desired vector space.
 A: I really don't like problems like this (this is a complaint for whoever assigned this problem, not you). "Can be made" is extremely vague, especially with no hypothesis on the field $F$. If $F$ is an infinite field like $\mathbb{R}$ or $\mathbb{C}$ then all you have to do is show that $U$ can be taken to have the same cardinality as $F$, since no compatibility is asked for between the vector space structure on $U$ and any other structure present, and every finite-dimensional vector space over an infinite field has the same cardinality. And if $F$ is a finite field $\mathbb{F}_q$ all you have to do is show that $U$ can be taken to have cardinality $\mathbb{F}_q^{r(n-r)}$.
Of course this is not the intended meaning. But context is required to discern the intended meaning and it's unfair to the student to not be more specific about what "can be made" means. For those in the know, $r(n - r)$ is famously the dimension of the Grassmannian $\text{Gr}_r(F^n)$ parameterizing $r$-dimensional subspaces of $F^n$, and in particular it is the dimension of the (Zariski, if $F$ isn't $\mathbb{R}$ or $\mathbb{C}$) tangent space at each point $W \in \text{Gr}_r(F^n)$. So the intended meaning, which is hard to discern with confidence without this extra context, is about linearly parameterizing the subspaces "near $W$" by $r(n - r)$ parameters.
Here's a sketch of how that goes. Pick a basis $w_1, \dots w_r$ of $W$, and imagine deforming that basis into a basis of a new subspace $w_1', \dots w_r' \in W'$. What kind of deformations can we perform? Well, consider just modifying just a single vector, say $w_1$, into a new vector $w_1' = w_1 + v_1$. What can $v_1$ be? If $v_1$ is just an element of $\text{span}(w_1, w_2, \dots w_n)$ we get the same vector space $W$ back (unless the component of $w_1$ in $v_1$ cancels $w_1$ in which case we get a proper subspace of $W$, which has the wrong dimension). More generally, whatever $v_1$ is, as long as the component of $w_1$ in it doesn't cancel $w_1$ out, only its image in $F^n/W$ affects the value of $\text{span}(w_1', w_2, \dots w_n)$. So $v_1$ effectively takes values in $F^n/W$, or said another way, if we pick a complement $V$ so that $F^n = W \oplus V$, then we might as well restrict our attention to $v_1 \in V$. This ensures, as Joppy mentions in the comments, that $\text{span}(w_1', w_2, \dots w_n)$ intersects $V$ trivially.
Now you can repeat this argument for each of the other vectors $w_i$, choosing $v_1, \dots v_r \in V$ to modify them by; note that $\dim V = n - r$ so the set of ways to do this naturally forms a vector space $\text{Hom}(W, V)$ of dimension $r(n - r)$. Then you can take $w_i' = w_i + v_i$ to be the basis of the new subspace $W'$, which still intersects $V$ trivially. It remains to check that different choices of the $v_i$ produce genuinely different subspaces. Once you do, you can take $U$ to be the set of all subspaces obtained in this way.
