Let $\alpha\in \Bbb{R}$ and consider the rotation $R_\alpha: \Bbb{R}/\Bbb{Z}\rightarrow \Bbb{R}/\Bbb{Z}$ a.t. $x\mapsto (x+\alpha)\operatorname{mod}\Bbb{Z}$. I want to check if $R_\alpha$ satisfies the shadowing property which is:
$\forall \epsilon >0$ there exists $\delta >0$ s.t. if $\{x_n\}_{n\in \Bbb{Z}}$ is a $\delta$-orbit for $R_\alpha$, then there exists $y\in \Bbb{R}/\Bbb{Z}$ s.t. $d(x_n, R_\alpha^n (y))<\epsilon$ for all $n \in \Bbb{Z}$.
On the internet I have found a similar exercise with a hint which told me that it does not satisfy the shadowing property. They wrote to consider $x_n:=n\alpha+n\gamma \operatorname{mod}\Bbb{Z}$ where $\gamma \in \Bbb{R}$ s.t. $|\gamma|<\delta$.
I could show that $\{x_n\}$ form a $\delta$-orbit for $R_\alpha$ and now I want to show with this that the shadowing property is not satisfied.
I thought that if I take an arbitrary $y\in \Bbb{R}/\Bbb{Z}$ say $y=y+\Bbb{Z}$ then $$\begin{align} d(x_n,R_\alpha^n (y))&=d(n(\alpha+\gamma)+\Bbb{Z}, y+n\gamma+\Bbb{Z})\\&=d(n\gamma,y)\stackrel{n\rightarrow \infty}{\longrightarrow }\infty\end{align}$$
Therefore I thought that it does not satisfy the shadowing property but does this work?
You are in the right direction, but I see a mistake as $d(n\gamma,y)\rightarrow\infty$ does not necessarily mean anything because we are actually dealing with the projections on the torus. For instance $d(n+\frac{1}{2},\frac{1}{2})\rightarrow\infty$ but on the torus they are the same.
What could work is the following: suppose by contradiction that the claim holds. Then for $\varepsilon = \frac{1}{10}$, there exists some $\delta>0$ (we can assume without loss of generality that $\delta<1$) so that for any $\delta$-orbit $x_n\in S^1$, there exists some $y\in S^1$, with $$d(x_n, R_\alpha^n y)< \frac{1}{10}$$ for all $n\in\mathbb{N}$.
For the $\delta$ given above, choose irrational $\gamma$ with $|\gamma|<\delta$ and let $\beta := \alpha-\gamma$ and $x_n = n\beta$.
Lemma 1: $x_n$ is a $\delta$-orbit. Proof: $d_{S^1}(R_\alpha x_n , x_{n+1}) = d_{S^1}(n\beta+\alpha, n\beta+\beta) = d_{S^1}(\alpha,\beta) = d_{S^1}(\gamma,0)< \delta,$ where $d_{S^1}$ is the distance in the torus (which is always bounded by the distance on $\mathbb{R}$).
Now to get a contradiction, it suffices to show that
Lemma 2: For every $y\in S^1$, there exists some $n$ so that $d(x_n,R_\alpha^n y)>\frac{1}{5}.$
Proof: First notice that we have $$d_{S^1}(n\alpha, y+n\beta) = d_{S^1}(n(\alpha-\beta), y) = d_{S^1}(n\gamma , y)$$
Since $\gamma$ is irrational, $n\gamma$ is dense in $S^1$. Let $y'\in S^1$ be the furthest point from $y$, so that $d(y,y') = \frac{1}{2}$. Therefore, we can find a subsequence $n_k$ so that $n_k\gamma\rightarrow y'$ in $S^1$.
Hence, we can find some $n_0$ so that $d_{S^1}(n_0\gamma,y')<\frac{1}{10}$.
By the triangle inequality $d_{S^1}(n_0\gamma,y) + d_{S^1}(n\gamma,y') > d_{S^1}(y,y')\Rightarrow d_{S^1}(n\gamma,y) = \frac{1}{2} - \frac{1}{10} > \frac{1}{5}$.
This completes the proof.