I encounter this problem in Noga Alon's book and I have been struggling to solve it: Prove that there exists a constant $C > 0$ such that for every $A \subset \mathbb{Z}_p$ where $|A| = k^2$, there exists $x \in \mathbb{Z}_p$ such that $S_x = \{xa: a \in A\}$ intersects with every interval length $Cp/k$.
The problem implies that there is some modulo class that is somewhat well-distributed, but I can't see how to apply the second-moment method here.
Let $xA := \{xa : a \in A\}$ and $xA+y := \{xa+y : a \in A\}$ for notational ease.
The starting point is to note that (i) there exists $x \in \mathbb{Z}_p$ such that $xA$ intersects every interval of length $\frac{Cp}{k}$ if and only if (ii) there exist $x,y \in \mathbb{Z}_p$ such that $xA+y$ intersects every interval of length $\frac{Cp}{k}$.
The benefit of analyzing $xA+y$ for random $x,y \in \mathbb{Z}_p$ instead of merely $xA$ for random $x \in \mathbb{Z}_p$ is that the former is more "uniform" than the latter, making the second moment method applicable.
Now for the (rigorous) argument. We prove (ii) above by the second moment method. In what follows, $x,y \in \mathbb{Z}_p$ will be independent random variables, with $x$ being uniformly chosen from $\mathbb{Z}_p\setminus\{0\}$ and $y$ being uniformly chosen from $\mathbb{Z}_p$.
Fix an interval $I \subseteq \mathbb{Z}_p$ of length $|I| = C\frac{p}{k}$. We have $$\underset{x,y \sim \mathbb{Z}_p}{\mathbb{E}}\bigg[\Big|I \cap (xA+y)\Big|\bigg] = \underset{x,y \sim \mathbb{Z}_p}{\mathbb{E}}\bigg[ \sum_{j \in I} 1_{j \in xA+y}\bigg] = \sum_{j \in I} \underset{x,y \sim \mathbb{Z}_p}{\mathbb{E}}\bigg[1_{j \in xA+y}\bigg] = \sum_{j \in I} \frac{|A|}{p} = \frac{|I|\,|A|}{p},$$ where the second to last equality holds due to the fact that any $x \in \mathbb{Z}_p\setminus\{0\}$ yields that $xA$ is a set of size $|A|$, so the probability over the choice of $y \in \mathbb{Z}_p$ that $xA+y$ contains $j$ is simply the probability that $y = j-xa$ for some $xa \in xA$, which is $\frac{|xA|}{p} = \frac{|A|}{p}$.
Similarly, we have $$\underset{x,y \sim \mathbb{Z}_p}{\mathbb{E}}\bigg[\Big|I \cap (xA+y)\Big|^2\bigg] = \underset{x,y \sim \mathbb{Z}_p}{\mathbb{E}}\bigg[ \Big(\sum_{j \in I} 1_{j \in xA+y}\Big)^2\bigg] = \underset{x,y \sim \mathbb{Z}_p}{\mathbb{E}}\bigg[\sum_{j_1,j_2 \in I} 1_{\substack{j_1 \in xA+y \\ j_2 \in xA+y}}\bigg] = \sum_{j_1,j_2 \in I} \underset{x,y \sim \mathbb{Z}_p}{\mathbb{E}}\bigg[1_{\substack{j_1 \in xA+y \\ j_2 \in xA+y}}\bigg].$$ Fix $j_1,j_2 \in I$. If $j_1 = j_2$, then $$\underset{x,y \sim \mathbb{Z}_p}{\mathbb{E}}\bigg[1_{\substack{j_1 \in xA+y \\ j_2 \in xA+y}}\bigg] = \underset{x,y \sim \mathbb{Z}_p}{\mathbb{E}}\bigg[1_{j_1 \in xA+y}\bigg] = \frac{|A|}{p},$$ as before. If $j_1 \not = j_2$, then $$\underset{x,y \sim \mathbb{Z}_p}{\mathbb{E}}\bigg[1_{\substack{j_1 \in xA+y \\ j_2 \in xA+y}}\bigg] = |A|(|A|-1)\frac{1}{(p-1)p},$$ since (1) for any distinct $a_1,a_2 \in A$, there exist unique $x \in \mathbb{Z}_p\setminus\{0\}, y \in \mathbb{Z}_p$ such that $xa_1+y = j_1$ and $xa_2+y = j_2$ and (2) if $j_1,j_2 \in xA+y$, then there exist unique (and distinct) $a_1,a_2 \in A$ with $j_1 = xa_1+y$ and $j_2 = xa_2+y$. Therefore, $$\underset{x,y \sim \mathbb{Z}_p}{\mathbb{E}}\bigg[\Big|I \cap (xA+y)\Big|^2\bigg] = \frac{|I|\, |A|}{p}+|I|(|I|-1)\frac{|A|(|A|-1)}{p(p-1)},$$ and consequently, \begin{align*} \underset{x,y \sim \mathbb{Z}_p}{\mathbb{E}}\Bigg[\bigg(\Big|I \cap (xA+y)\Big|-\frac{|I|\, |A|}{p}\bigg)^2\Bigg] &= \frac{|I|\, |A|}{p}+|I|(|I|-1)\frac{|A|(|A|-1)}{p(p-1)} - \frac{|I|^2\,|A|^2}{p^2} \\ &= \frac{|I|\, |A|}{p}\bigg[1+(|I|-1)\frac{|A|-1}{p-1}-\frac{|I|\, |A|}{p}\bigg] \\ &= \frac{|I|\, |A|}{p}\frac{(p-|A|)(p-|I|)}{p(p-1)}. \end{align*}
Now we use the second moment method: for any $C' > 0$, we have \begin{align*}\underset{x,y \sim \mathbb{Z}_p}{\mathbb{P}}\Bigg[\bigg|\big|I \cap (xA+y)\big|-\frac{|I|\, |A|}{p}\bigg| > C'\sqrt{\frac{|I|\, |A|}{p}}\Bigg] &= \underset{x,y \sim \mathbb{Z}_p}{\mathbb{P}}\Bigg[\bigg|\big|I \cap (xA+y)\big|-\frac{|I|\, |A|}{p}\bigg|^2 > (C')^2\frac{|I|\, |A|}{p}\Bigg] \\ &\le \frac{1}{(C')^2\frac{|I|\, |A|}{p}}\frac{|I|\,|A|}{p} \frac{(p-|A|)(p-|I|)}{p(p-1)} \\ &= \frac{(p-|A|)(p-|I|)}{(C')^2 p(p-1)} \\ &\le \frac{1}{(C')^2}. \end{align*}
Letting $C' = \sqrt{k}$ gives $$\underset{x,y \sim \mathbb{Z}_p}{\mathbb{P}}\Bigg[\bigg|\big|I \cap (xA+y)\big|-Ck\bigg| > \sqrt{C}k\Bigg] \le \frac{1}{k},$$ which in particular means $$\tag{1}\label{intbound} \underset{x,y \sim \mathbb{Z}_p}{\mathbb{P}}\Bigg[I \cap (xA+y) = \emptyset \Bigg] \le \frac{1}{k}.$$
We're almost done. Take $m := 2\frac{k}{C}$ disjoint intervals in $\mathbb{Z}_p$, each of length $\frac{1}{2}\frac{Cp}{k}$; call them $I_1,\dots,I_m$. Since \eqref{intbound} holds for each $I_j$, namely $$ \underset{x,y \sim \mathbb{Z}_p}{\mathbb{P}}\Bigg[I_j \cap (xA+y) = \emptyset \Bigg] \le \frac{1}{k},$$ by a union bound we have $$\underset{x,y \sim \mathbb{Z}_p}{\mathbb{P}}\Bigg[\exists 1 \le j \le m \text{ such that } I_j \cap (xA+y) = \emptyset \Bigg] \le m\frac{1}{k} = \frac{2}{C}.$$ In particular, if $C > 2$ (say $C=3$), we have $$\underset{x,y \sim \mathbb{Z}_p}{\mathbb{P}}\Bigg[\exists 1 \le j \le m \text{ such that } I_j \cap (xA+y) = \emptyset \Bigg] < 1.$$ Hence, (for $C > 2$), there exist $x,y \in \mathbb{Z}_p, x \not = 0$ such that $xA+y$ intersects each of $I_1,\dots,I_m$. Let $x_*,y_*$ denote such $x,y$.
To finish, take an arbitrary interval $I \subseteq \mathbb{Z}_p$ of length $|I| = C\frac{p}{k}$. As noted at the beginning of this answer, it suffices to show $x_*A+y_*$ intersects $I$. The critical thing to note is that $I \supseteq I_j$ for some $1 \le j \le m$. Indeed, this follows from the fact that each $|I_j| \le \frac{1}{2}|I|$, so that if $j$ is minimal with respect to $I_j \cap I \not = \emptyset$, then ($j \le m-1$ and) $I \supseteq I_{j+1}$. Since $x_*A+y_*$ intersects $I_j$ by assumption, it follows that $x_*A+y_*$ intersects $I$, as desired. $\square$
Notes
The answer here to the same question is invalid, since it incorrectly appeals to independence.
The answer here to the same question is invalid (though close to valid), since it only proves that (in our notation) $xA+y$ intersects $\frac{k}{C}$ disjoint intervals each of length $C\frac{p}{k}$, but doesn't prove $xA+y$ intersects all intervals of length $C\frac{p}{k}$ (we handle this by introducing a factor of $2$).