Let$\{X_n\}_{n=1}^{\infty}$ be a sequence of i.i.d. random variable. Let $\{\alpha_k\}_{k=1}^{\infty} $be a sequence of
strictly increasing finite stopping times.
Then $\{X_{\alpha_k+1}\}_{n=1}^{\infty}$ is also a sequence of i.i.d. random variable.
I can prove this theorem under the assumption that the family $\{X_n,\alpha_k\}_{n,k\in\mathbb{N}}$ is independent.
Actually, to prove the independence I just prove the case of finite intersection and the independence of infinite intersection follows by taking the limit. Here is my proof:
Let $\Lambda_n=\bigcap\limits_{k=1}^{n} \{X_{\alpha_k+1} \in B_k\} $ for $ n\in\mathbb{N}$
Claim: $P(\Lambda_n)=\prod\limits_{k=1}^{n}P(X_{k} \in B_k)$
$$P(\Lambda_1)=\sum\limits_{m=0}^{\infty}P(X_{m+1} \in B_1,\alpha_1=m) \\
=\sum\limits_{m=0}^{\infty}P(X_{m+1} \in B_1)P(\alpha_1=m) \\
=\sum\limits_{m=0}^{\infty}P(X_{1} \in B_1)P(\alpha_1=m)\\=P(X_{1} \in B_1)$$
Assume it is true for $n=N-1$.
$$P(\Lambda_N)=\sum\limits_{m=N}^{\infty}P(X_{m+1} \in B_N,\alpha_N=m,\Lambda_{N-1})\\
=\sum\limits_{m=N}^{\infty}\sum\limits_{0\leq i_1 < i_2<...<i_{N-1}<m}P(X_{m+1} \in B_N,\alpha_N=m,\alpha_k=i_k,X_{i_k+1}\in B_k , k=1,2,...,N-1) \\
=\sum\limits_{m=N}^{\infty}\sum\limits_{0\leq i_1 < i_2<...<i_{N-1}<m}P(X_{m+1} \in B_N)P(\alpha_N=m,\alpha_k=i_k,X_{i_k+1}\in B_k , k=1,2,...,N-1) \\
=P(X_N\in B_N)P(\Lambda_{N-1})$$
As $P(X_{\alpha_k+1}\in B)=P(X_{k} \in B)$ (from the proof for n=1),
the independence is proved.
And by the same reason, $P(X_{\alpha_k+1}\in B)=P(X_{k} \in B)=P(X_{1} \in B)=P(X_{\alpha_1+1}\in B)$ for all $k\in \mathbb{N}$, it is identically distributed.
I am not sure whether my proof is correct.
The main problem is whether assumption "The family $\{X_n,\alpha_k\}_{n,k\in\mathbb{N}}$ is independent." can be dropped. Since one can take $\alpha_k $ to be the first hitting time of ${X_k} $ with some set, it is not suitable to assume this family is independent.
If yes, how to prove the theorem without this assumption?
If no, is there any counter example?
If the sequence $\{X_n\}$ is i.i.d., and if the $\alpha_k$ are stopping times of the filtration $\mathcal F_n:=\sigma\{ X_k: k=1,2,\ldots,n\}$, $n\ge 1$, then your argument as it stands shows that $\{X_{\alpha_k+1}\}_{k=1}^\infty$ is i.i.d. For example, in considering $$ P(X_{m+1} \in B_N,\alpha_N=m,\Lambda_{N-1}) $$ in your calculation, for $m\ge N$, the event $\{\alpha_N=m\}\cap \Lambda_{N-1}$ is an element of $\mathcal F_m$, hence independent of $X_{m+1}$. Therefore $$ P(X_{m+1} \in B_N,\alpha_N=m,\Lambda_{N-1})=P(X_{m+1} \in B_N)\cdot P(\alpha_N=m,\Lambda_{N-1}), $$ as required.