Let $X_n$ be a sequence of uniformly bounded random variables — that is, there exists some $K > 0$ such that $|X_n| \leq K$ almost surely for all $n \in \mathbb N$.
Write $\bar X_N := \frac{1}{N} \sum_{n = 1}^N X_n$, and $Y := \limsup_{N \to \infty} \bar X_N$.
Question: Does there exist some subsequence $X_{n_k}$ such that $\frac{1}{N} \sum_{k=1}^{N} X_{n_k}$ converges to $Y$ almost surely as $N \to \infty$?
No I do not think this is always possible. Take for example a subsequence of $\mathbb{N}$, call it $\{ m_k \}$ and consider the random variables \begin{equation} X_i = \chi_{(0,1/2)},\quad m_k \leq i < m_{k+1} \end{equation} if $k$ is even and let $X_i = \chi_{(1/2,1)}$ otherwise. Then if $m_k$ is growing sufficiently fast we have $Y \equiv 1$ (with your notation). So suppose that for some subsequence $n_k$ you have that $\frac{1}{N}\sum_{n=1}^{N} X_{n_k} =:Z_k $ converges almost everywhere to $1$, by Fatou's lemma you would have \begin{equation*} 1 = \int_0^1 \liminf Z_k \leq \liminf_k \int_0^1 Z_k = \frac{1}{2}. \end{equation*}
