Let $(T_n)$ be a sequence of stopping times. If $\limsup_{n \to \infty} T_n$ is finite, then it is a stopping time I'm doing this exercise from my lecture note of martingale theory:

Let $I \subseteq \mathbb N$ and $(\mathcal F_n)$ be a filtration. A random variable $T$ taking values in $I \cup \{\infty\}$ is a stopping time if and only if $\{T = n\} \in \mathcal F_n$ for all $n \in I$.
$\textbf{Proposition:}$ Let $(T_n)$ be a sequence of stopping times. If $T = \limsup_{n \to \infty} T_n$ is finite, then $T$ is a stopping time. Similarly, If $T = \liminf_{n \to \infty} T_n$ is finite, then $T$ is a stopping time.

Could you please verify if my proof is correct or contains logical mistake? Thank you so much!

My attempt:
We need two lemmas:

$\textbf{Lemma 1:}$ If $(T_n)$ is a sequence of stopping times such that $T = \lim_{n \to \infty} T_n$ exists and is finite. Then $T$ is a stopping time.
$\textbf{Lemma 2:}$ If $S,T$ are stopping time, then so are $S \vee T$ and $S\wedge T$.

First, we prove that $\inf_{k \ge N} T_k$ is a stopping time for any $N \in \mathbb N$. Define a sequence $(X_n)$ by $X_0 = T_N$ and $X_{n+1} = X_n \wedge T_{N+n+1}$. By $\textbf{Lemma 2}$, $(X_n)$ is a sequence of stopping times such that $\lim_{n \to \infty} X_n = \inf_{k \ge N} T_k$. Then, by $\textbf{Lemma 1}$, $\inf_{k \ge N} T_k$ is a stopping time.
On the other hand, $T = \limsup_{n \to \infty} T_n = \lim_{n \to \infty} \inf_{k \ge n} T_k$. By $\textbf{Lemma 1}$ again, $T$ is a stopping time.
In a similar manner, $\liminf_{n \to \infty} T_n$ is a stopping time.
 A: From @TheBridge, I supplement my original proof with details here.

My attempt: We need two lemmas:

$\textbf{Lemma 1:}$ If $(T_n)$ is a sequence of stopping times such that $T = \lim_{n \to \infty} T_n$ exists and is finite. Then $T$ is a stopping time.
$\textbf{Lemma 2:}$ If $(S_n)_{m=0}^n$ is a finite sequence of stopping times, then $\max_{0 \le m \le n} S_m$ and $\min_{0 \le m \le n} S_m$ are stopping time.

First, we prove that $\inf_{m \ge n} T_m$ is a stopping time for any $n \in \mathbb N$. It suffices to prove that $\inf_{m \ge 0} T_m$ is a stopping time. Define a sequence $(X_n)$ by $X_{n} = \min_{0 \le m \le n} T_m$. Clearly, $\inf_{m \ge 0} T_m \le X_{n}$. Given $(\epsilon, \omega) \in \mathbb R^+ \times \Omega$, there is $k \in \mathbb N$ such that $\inf_{m \ge 0} T_m (\omega) \le T_k (\omega) < \inf_{m \ge 0} T_m (\omega)+ \epsilon$. Then $\inf_{m \ge 0} T_m (\omega) \le X_k (\omega) < \inf_{m \ge 0} T_m (\omega)+ \epsilon$. It follows that $\lim_{n \to \infty} X_n(\omega) = \inf_{m \ge 0} T_m (\omega)$ and hence $\lim_{n \to \infty} X_n = \inf_{m \ge 0} T_m$. By $\textbf{Lemma 2}$, $(X_n)$ is a sequence of stopping times. Then, by $\textbf{Lemma 1}$, $\lim_{n \to \infty} X_n$ is also a stopping time. Hence $\inf_{m \ge 0} T_m$ is a stopping time.
On the other hand, $T = \liminf_{n \to \infty} T_n = \lim_{n \to \infty} \inf_{m \ge n} T_m$. By $\textbf{Lemma 1}$ again, $T$ is a stopping time. In a similar manner, $\limsup_{n \to \infty} T_n$ is a stopping time.
A: Here is how I see it (using your lemmas 1 and 2)
Let's look at the "$\liminf$" case :
By hypothesis(and wiki page section 1) :
$$\liminf _n T_n =\sup_n \inf_{k>n} T_k=T$$ exists and is finite. 
So for some $n_0$ big enough we have for all $n>n_0$:
$$I_n =\inf_{k>n} T_k$$ is finite. 
Then we can use your line of argument to prove that $I_n$ is a stopping time. for every such $n$, but please pay attention to the indexation. 
Defining  $I^p_n = T_n\wedge ... \wedge T_p$ and using lemma 2, $I^p_n$ is a stopping time for every $p$. As $\lim_p I^p_n= I_n$ is finite, we are now ok to use lemma 1, and conclude that $I_n$ is a stopping time. 
Let's finish the job. You can observe that $T= \sup_n I_n = \lim_n I_n$ (observe that $I_n$ is increasing) exists and is finite where we know that $I_n$ is a sequence of stopping times.(we can restart the sequence $I_n$ at n_0 if needed). So we can  re-use lemma 1 directly to conclude that $T$ is a stopping time. 
The "$\limsup$" case is treated symmetrically.
