I was reading lecture on martingale where the following result is proved. Unfortunately it's not very detailed.
I appreciate if you could explain why $Y_\infty^{(p)}$ are equal $\mathbb{P}$-a.s, from the fact $Q_{p+1}\subset Q_p$ and how to conclude from the right continuity that $X_r \to Y_{\infty} \ \mathbb{P}$-a.s.
Update: perhaps the a.s equality is true because, $Y_{k}^{(p)}=Y_{2k}^{(p+1)},$ the limit is unique a.s so $Y_{\infty}^{(p)}=Y_{\infty}^{(p+1)},$ for all $p \geq 1$, so $Q_p$ are not necessary.
Your reason for why $Y^{(p)}_\infty = Y^{(p+1)}_\infty$ is correct, but I'm not sure I agree with the statement that "$Q_p$ are not necessary." The reason we have $Y_{2k}^{(p+1)} = Y_{k}^{(p)}$ is because of the fact that $Q_{p+1} \subset Q_p$, so it is necessary we use something like $Q_p$ here.
Now to conclude $X_r \rightarrow Y_\infty$, note that by right-continuity of $X$ we have $$\limsup_{r \rightarrow \infty} X_r = \limsup_{r \rightarrow \infty, r \in Q} X_r$$ where $Q := \bigcup_{p \in \mathbb{N}} Q_p$ and similarly $$\liminf_{r \rightarrow \infty} X_r = \liminf_{r \rightarrow \infty, r \in Q} X_r.$$
However, because $Y_k^{(p)} \rightarrow Y_\infty$ a.s., we know $\limsup_{r \rightarrow \infty, r \in Q} X_r = Y_\infty$ and $\liminf_{r \rightarrow \infty, r \in Q} X_r = Y_\infty$ a.s. Since $$\limsup_{r \rightarrow \infty} X_r = \liminf_{r \rightarrow \infty} X_r = Y_\infty$$ we conclude $\lim_{r \rightarrow \infty} X_r = Y_\infty$.
EDIT: After thinking about this more carefully, I don't think this is correct, and I don't think the proof presented in the original question works either for the same reason. This doesn't use the fact that $X$ is a sub-martingale except to conclude each $Y_k^{(p)}$ converges, but we could make a right-continuous process that doesn't converge, and still has the property that each $Y_k^{(p)}$ converges. For example, the process that is $1$ on intervals of the form $[n+2^{-n},n+2^{1-n})$ and $0$ everywhere else, i.e. \begin{align*} X_r := \sum_{n=1}^\infty 1_{[n+2^{-n},n+2^{1-n})}(r), \end{align*} has the property that each $Y_k^{(p)}$ is eventually constant at $0$, but $\limsup_{r \rightarrow \infty} X_r = 1$.
The typical way I've seen this proof is to look at downcrossings and use the downcrossing inequality, e.g. Section 2.2 in Continuous Martingales and Brownian Motion by Revuz and Yor. Here's a sketch of that proof:
For a countable subset $E \subset \mathbb{R}$ and $a < b$, let $D(a,b,E)$ be the number of downcrossings of the interval $(a,b)$ by $X|_E$ ($X$ restricted to $E$). The discrete time downcrossing inequality tells us $D(a,b,E)$ is finite almost surely for all countable $E$, so by choosing $E$ to be a countable dense subset of $\mathbb{R}$ and using right-continuity we conclude that $X$ can only have finitely many downcrossings in any interval $(a,b)$. This implies $\liminf_{r \rightarrow \infty} X_r = \limsup_{r \rightarrow \infty} X_r$ because if $\liminf_{r \rightarrow \infty} X_r < \limsup_{r \rightarrow \infty} X_r$ then there must exist $a,b \in \mathbb{Q}$ such that $\liminf_{r \rightarrow \infty} X_r < a < b < \limsup_{r \rightarrow \infty} X_r$, but that implies $X_r$ has infinitely many downcrossings in the interval $(a,b)$ and therefore $P(\liminf_{r \rightarrow \infty} X_r < \limsup_{r \rightarrow \infty} X_r) = 0$.