Let $X \subseteq \mathbb{R}$ be a compact set. Show that $\mathcal{E} = \{e^{int}\}_{n \in \mathbb{Z}}$ is a Riesz sequence in $L^2(X)$ if and only if $X+2\pi \mathbb{Z} = \mathbb{R}$.
We say that $\mathcal{E}$ is a Riesz sequence in $L^2(X)$ if we have the following chain of inequalities: $$A \sum_n |c_n|^2 \leq \left\|\sum_n c_n e^{int}\right\|_{L^2(X)}^2 \leq B \sum|c_n|^2,$$ where $c = (c_n)$ is a finite sequence.
Suppose that $X+2\pi\mathbb{Z} = \mathbb{R}$. Then since $X$ is bounded there is $N \in \mathbb{N}$ such that $X \subseteq [-\pi N, \pi N]$. With this we get the the latter inequality by using the fact that $\mathcal{E}$ forms an orthogonal basis in $L^2[-\pi,\pi]$ and the periodicity of its elements. However I am unsure how to achieve the fitst inequality. One idea I had (which may be wrong) is to cover the interval $[-\pi,\pi]$ by a finite number of translates of the form $\{X+2\pi k\}_{k=1}^N$. However, I cannot convince myself that I can do this only using a finite number of translates. If this is true, then this direction is done. Otherwise, I am unsure. As for the other direction, I have no clue on how to start. Of course one only needs to show that every real number $r \in \mathbb{R}$ can be written in the form $r = x+2\pi m$ for some $m \in \mathbb{Z}$. Any help is appreciated, thanks!
This question is best visualized in the circle, so let me first reduce it to that situation.
Let us therefore consider the projection $q$ from ${\mathbb R}$ to $S^1$ given by $$ q:t\in {\mathbb R}\mapsto e^{it}\in S^1, $$ and let us put $Z=q(X)$. The condition that $X+2\pi \mathbb{Z} = \mathbb{R}$ is therefore clearly equivalent to $Z=S^1$.
On the other hand, in order to rephrase the Riesz condition, I claim that the following are equivalent:
(The stated Riesz condition on ${\mathbb R}$). There exist positive real numbers $A$ and $B$ such that, for all sequences of complex numbers $\{c_n\}_{n\in {\mathbb N}}$, with finitely many nonzero terms, one has that $$ A \sum_n|c_n|^2 \leq \underbrace{\Big\|\sum_n c_n e^{int}\Big\|_{L^2(X)}^2}_{I_1} \leq B \sum_n|c_n|^2. \tag {$*$} $$
(The Riesz condition on $S^1$). There exist positive real numbers $A$ and $B$ such that, for all sequences of complex numbers $\{c_n\}_{n\in {\mathbb N}}$, with finitely many nonzero terms, one has that $$ A \sum_n|c_n|^2 \leq \underbrace {\Big\|\sum_n c_n z^n\Big\|_{L^2(Z)}^2}_{I_2} \leq B \sum_n|c_n|^2, \tag {$**$} $$ where $z^n$ denotes the $n^{th}$ function in the standard orthonormal basis for $L^2(S^1)$, namely $z\mapsto z^n$.
In order to prove that (1) $\Leftrightarrow$ (2), and putting $$ X_n=X\cap [2\pi n, 2\pi (n+1)), $$ observe that $ X=\bigcup_{n=-N}^N X_n, $ for some $N$.
So if we are given $\{c_n\}_{n\in {\mathbb N}}$, as above, and we set $f(z)=\sum_n c_n z^n$, we have $$ \Big\|\sum_n c_n e^{int}\Big\|_{L^2(X)}^2 = \frac 1{2\pi }\int_X|f(e^{it})|^2\,dt= $$$$= \frac 1{2\pi }\sum_{n=-N}^N \int_{X_n} |f(e^{it})|^2\,dt= \sum_{n=-N}^N \int_{q(X_n)} |f(z)|^2\,dz, $$ where the last integral is taken with respect to the normalized Haar measure on the circle.
Regarding the last term displayed above, and noticing that $Z=\bigcup_{n=-N}^N q(X_n)$, (incidentally no longer a disjoint union), we have $$ \int_{Z} |f(z)|^2\,dz \leq \sum_{n=-N}^N \int_{q(X_n)} |f(z)|^2\,dz \leq $$$$ \leq \sum_{n=-N}^N \int_{Z} |f(z)|^2\,dz = (2N+1) \int_{Z} |f(z)|^2\,dz. $$
Summarizing, if we denote the middle term in ($*$) by $I_1$, and the middle term in ($**$) by $I_2$, we have shown that $$ I_2\leq I_1\leq (2N+1)I_2. $$ This said it is now clear that (1) $\Leftrightarrow$ (2).
(EDIT: The remaining part of this answer was reformulated, as follows.)
With the appropriate translations provided by the work done so far, the equivalent "circle version" of the present question becomes:
Theorem. If $Z\subseteq S^1$ is a closed subset, then the following are equivalent:
$Z=S^1$,
(2) holds.
Proof. The forward implication clearly follows from Parseval's identity (with $A=B=1$). For the reverse direction, consider the orthogonal projection $P$ from $L^2(S^1)$ to $L^2(Z)$, namely the map $$ P: f \mapsto f1_Z, $$ where $1_Z$ is the characteristic function of $Z$. Given a sequence of complex numbers $\{c_n\}_{n\in {\mathbb N}}$ with finitely many nonzero terms, let $f$ be the function on the circle given by $$ f(z)=\sum_n c_nz^n, $$ and observe that $$ \|f\|_{L^2(S^1)}^2 = \sum_n|c_n|^2, $$ while $$ \|P(f)\|_{L^2(S^1)}^2 = \|f\|_{L^2(Z)}^2 = \Big\|\sum_n c_n z^n\Big\|_{L^2(Z)}^2. $$ Therefore we deduce from (2) that $$ A \|f\|^2 \leq \|P(f)\|^2 \leq B \|f\|^2, \tag {$\dagger$} $$ where all norms are from $L^2(S^1)$. Observing that the functions $f$ so far considered form a dense subspace of $L^2(S^1)$, we conclude that $(\dagger$) holds for every $f$ in $L^2(S^1)$.
In particular we see that $P$ is one-to-one. On the other hand, the kernel of a projection is known to be the space orthogonal to its range, whence $$ \{0\} = \text{Ker}(P) = L^2(Z)^\perp = L^2(S^1\setminus Z). $$ So necessarily $S^1\setminus Z$ has measure zero, but since this is an open set, we deduce that $S^1\setminus Z$ is empty, and hence $Z=S^1$, as desired. QED