Let $(x_n)$ be a real sequence and $\beta_n$ be any strictly increasing sequence of positive integers.
We say that $x_n\in D_{so}$, if for all $\epsilon>0$ there exists $\delta=\delta(\epsilon)$ and $N=N(\epsilon)$ such that $\mid x_m-x_{\beta_n} \mid<\epsilon$ whenever $n\geq N(\epsilon)$ and $\beta_n< m\leq (1+\delta)\beta_n$.
I want to prove: If $x_n\in D_{so}$ and $(x_{\beta_n})$ is convergent to $L$, then $(x_n)$ is convergent to $L$.
I tried:
Firstly assume that $(x_{\beta_n})$ is convergent to $L$. Then for all $\epsilon>0$ there exists and $N_0=N_0(\epsilon)$ such that $\mid x_{\beta_n}-L \mid<\epsilon$ whenever $n\geq N_0(\epsilon)$.
Now, assume $x_n\in D_{so}$. Then for all $\epsilon>0$ there exists $\delta=\delta(\epsilon)$ and $N=N(\epsilon)$ such that $\mid x_m-x_{\beta_n} \mid<\epsilon$ whenever $n\geq N(\epsilon)$ and $\beta_n< m\leq (1+\delta)\beta_n$.
Put $M=\max\{N,N_0\}$. Then, for all $n\geq M$ and accordingly for all $m,\beta_n\geq M$ (since $\beta_n$ is as large as $n$), we have $$\mid x_m-L \mid \leq \mid x_m-x_{\beta_n} \mid + \mid x_{\beta_n}-L \mid<\epsilon+\epsilon=2\epsilon.$$ Therefore, $(x_n)$ is convergent to $L$.
This needn't be true.
Let $\beta_n=2^n$ and $x_n=0$ if $2^m\leq n\leq 1.5\times 2^m$ for some $m\geq 1$, $x_n=1$ otherwise.
Then note that for all $\epsilon >0$, with $\delta(\epsilon)=0.5$ and $N(\epsilon)=1$, if $n\geq N$, and $\beta_n\leq m\leq (1+\delta)\beta_n$, then by construction of the sequence, $|x_m-x_{\beta_n}|=|0-0|=0<\epsilon$. So $x_n\in D_{so}$ . Moreover, clearly $x_{\beta_n}=0$ for all $n$, so in particular $(x_{\beta_n})$ converges to $0$.
However, note that for all $n\geq 2$, $1.5\times 2^n = 2^n + 2^{n-1} < 2^n + 2^n - 1 = 2^{n+1}-1< 2^{n+1}$. So for all $n\geq 2$, $x_{2^{n+1}-1}=1$.
So $(x_n)$ does not converge to $0$.