Why should it be hard to generalize Dvir's proof of the finite field Kakeya conjecture to the Euclidean case? Let $q$ be prime and let
$q\delta \sim 1.$ Let $K$ be any set of $C_n\delta$-separated tubes in $B(0,2)$, where $C_n$ is some constant depending on $n$. Let us consider a grid of $q^n$ points scaled with $\delta^n$. We embed this evenly spaced orthogonal grid in $\mathbf{R}^n$ so that the origins of the two sets meet. Let $K' \subset K$ so that each tube in $K$ contains a line from $K'$. The lines  $a + vt \in K'$ are such that if $v \in tu$ then $a + tu = a + vt$. So that we have $q^{n−1}$ lines in $K$ and the directions are $\delta$-separated (edit: not necessarily true if $K' \subset K$) . By Dvir's theorem we have $\lvert K'\rvert \ge \frac{q^n}{2^n}$. So we have at least that many points $x \in K' \subset K$ that are $\delta$-separated. Multiplying above by $\delta^n$ we have $\lvert K\rvert  \ge \lvert K'\rvert\delta^n \ge q^n\delta^n \frac{1}{2^n} \gtrsim \frac{1}{2^n}$ , because the $C_n\delta$-tubes contain the $\delta$−tubes. Isn't this a straightforward generalization? (EDIT) As was pointed out, those "grid lines" are unlike euclidean lines. The grid lines can go to the end of the grid and then reappear somewhere else in the grid. If a grid line contains $m$ points, it seems to me that for those "grid lines" we may even need  (in euclidean sense) $m$ parallel lines to the cover the grid line. This grows the number of (euclidean) lines and solves the mystery (for me). The solution: Let $K$ contain in addition to $\delta$-separated tubes essentially parallel tubes in order to achieve $K' \subset K.$ However, estimating how many parallel tubes you need may be difficult.
 A: I don't entirely follow your proposed approach, but let me make a few remarks which might help you think about this:
(1) Clearly this can't work, since this argument purports to show that a Kakeya set must have positive Lebesgue measure which, in contrast to the finite field setting, we know is false in Euclidean space.
(2) Note that the discretization of a tube is not necessary an algebraic line or even approximately an algebraic line. Indeed consider R^2 and a tube oriented along the y axis. Now adjust the angle slightly in either direction. What does a discretization of this tube look like? Well it depends on the scales involved but, roughly in the case of a $\delta$ tube rotated by a $~\delta$ degree angle, the discretization will be the union of long segments of vertical lines (rectangles/blocks) stacked on top of each other slowly drifting in the direction of the rotation. This is not an algebraic line.
(3) In the discrete (finite field) setting two lines intersect at a single point. However, for reasons similar to the discussion in (2), two tubes can have substantial overlap. This is sometimes refereed to as "small angle issues" in the literature.
