Your proof of 3.17 makes many assumptions that are not obvious to me.
Here is my proof of it for anyone who stumbles across this post and is similarly confused about 3.17.
3.17(a) case 1:
$s^* = +\infty$ imples that $E$ is not bounded above (suppose not, then we would have that sup $E < +\infty$).
$E$ is not bounded above implies that $\{s_n\}$ is not bounded above. To see why, suppose that $E$ is not bounded above, but $\{s_n\}$ is. Let $x$ = sup $\{s_n\}$. Since $E$ is not bounded above, there exists some $y \in E$ such that $y > x$. Thus, for some $\{s_{n_k}\}$, for all $\epsilon > 0$, there is some $N$ such that for all $n_k \ge N$, we have that $d(y, s_{n_k}) < \epsilon$.
Specifically, set $\epsilon = d(y, s)$. Then there exists an $s_{n_k}$ such that $d(y, s_{n_k}) < \epsilon$, which implies that $d(y, s_{n_k}) < d(y, x)$. We know that $y > x > s_{n_k}$ (by assumption and by the definition of $x$).
Therefore:
$d(y, s_{n_k}) < d(y, x)$
$\Rightarrow y - s_{n_k} < y - x$
$\Rightarrow s_{n_k} > x$, which contradicts the fact that $x =$ sup $\{s_n\}$.
Hence, we can conclude that $\{s_n\}$ is not bounded above!
We now construct a subsequence of $\{s_n\}$ as follows:
Let $s_{n_k} > k$ for all $k \in \mathbb{Z}^+$, with $n_1 < n_2 < n_3 < \dots$.
Suppose that we could not construct such a subsequence. If there were no $n_1$ such that $s_{n_1} \ge 1$, then $\{s_n\}$ would be bounded above (by 1), a contradiction. If there were no $n_k > n_i$ for $i \in \{1, \dots, k-1\}$ such that $s_{n_k} \ge k$, then we would have that $s_n < k$ for all $n > n_{k-1}$.
Thus, $\{s_n\}$ would be bounded by max($\{s_1, s_2, \dots, s_{n-k}\},k$), which contradicts the fact that $\{s_n\}$ is not bounded.
But this subsequence guarantees that for all $M$, there exists an $M$ such that there is some $N$ such that for all $n_k \ge N$, $s_{n_k} \ge M$. Hence $s_{n_k} \rightarrow +\infty$. Thus $+\infty \in E$.
3.17(a) case 2:
Now suppose $s^* \in \mathbb{R}$. Then $E$ is bounded above (by $s^*$). We now prove that this implies that at least one subsequential limit exists.
Suppose not. That is, suppose that $E$ were bounded above by $s^*$ and no subsequential limit existed. If no such limit existed, then there would be no $s \in \mathbb{R}$ such that $s \in E$. Hence, $E$ could only be four possible things: $\{+\infty, -\infty\}, \{+\infty\}, \{-\infty\}, \emptyset$. But we know that $+\infty \not \in E$, because if it were in $E$, that would contradict the fact that $s^* =$ sup $E$ is in $\mathbb{R}$.
Now suppose that either $E = \{-\infty\}$ or $E = \emptyset$. Then we would have that for any $y \in \mathbb{R}$ such that $y < s^*$, for all $x \in E$, $x < y$ (note this is obviously true when $E$ consists solely of $-\infty$ and is vacuously true when $E$ is the empty set). Hence we would have an upper bound of $E$ that is less than $s^*$ which contradicts the fact that $s^* = $ sup $E$. Thus, there is a subsequential limit of $\{s_n\}$.
By 3.7, we know that the subsequential limits of any sequence in metric space $X$ form a closed subset of $X$.
By 2.28, we know that for any non-empty subset $E$ of the real numbers that is bounded above, sup $E \in \overline{E}$.
Hence $s^* \in E$.
3.17(a) case 3:
Suppose that $-\infty =$ sup E. Then we know that $+\infty \not \in E$, and that $x \not \in E$ for all $x \in \mathbb{R}$.
We will now show that for every $M$, there exists an $N$ such that for all $n \ge N$, $s_n \le M$. This suffices to show that $s_n \rightarrow -\infty$, and hence that $-\infty \in E$.
Suppose not. That is, suppose that there is some $M$ such that for all $N$, there is some $n \ge N$ such that $s_n > M$. This implies that $s_n > M$ for infinitely many values of $s_n$. Let us consider the subsequence $\{s_{n_k}\}$ consisting of these values.
We know that $\{s_{n_k}\}$ is bounded below by $M$. There are two possibilities:
(I) $\{s_{n_k}\}$ is not bounded above. Hence, by the argument we made in 3.17(a) case 1, $+\infty \in E$, which contradicts the fact that $-\infty = s^* = $ sup $E$.
(II) $\{s_{n_k}\}$ is bounded above. Then $\{s_{n_k}\}$ is bounded below by $M$ and above by $\beta$, for some $\beta$. Hence, $\{s_{n_k}\}$ is bounded. But by 3.6(b) every bounded sequence in $\mathbb{R}$ contains a convergent subsequence. Hence, some subsequence of $\{s_n\}$ converges to some point $p \in \mathbb{R}$, which contradicts the fact that $-\infty = s^* =$ sup $E$.
Thus, for all $M$, $s_n > M$ for only finitely many values of $s_n$. Thus $s_n \rightarrow -\infty$. Therefore, $s^* \in E$.
Your reasoning seems right to me, but conceptually speaking I believe the proof is much more easy.
Proof from Rudin's definitions:
Put $t^*=\limsup_{n\to\infty} t_n$ and $s^*=\limsup_{n\to\infty}s_n$. For a contradiction, suppose that $s^*>t^*$ and put $\epsilon=\frac{s^*-t^*}{2}, x=s^*-\epsilon$. Notice that $x>t^*$.
With the definitions as in Rudin's book, by Theorem 3.17(2), there is $N_0\in\mathbb{N}$ such that for all $n\geq N_0$ we have $t_n<x$. On the other hand, by Theorem 3.17(1), $s^*\in E$ and so there is a subsequence $s_{n_k}$ that converges to $s^*$. The last implies in particular that there is $n_k\geq \max\{N_0,N\}$ such that $s_{n_k}\in (s^*-\epsilon,s^*+\epsilon)$, and we would have $$s_{n_k}>s^*-\epsilon=x>t_{n_k},$$ contradicting that $s_n\leq t_n$ for $n\geq N$.
To prove that $\liminf_{n\to\infty} s_n\leq \liminf_{n\to\infty}t_n$ you have two options: either you proof the analog of Theorem 3.17 for $s_*$, or you show that for a sequence $\{x_n\}$, $$\liminf_{n\to\infty} x_n=\limsup_{n\to\infty} (-x_n).$$
Proof with other definition:
First, note that the following:
Claim: If $\{s_n\},\{t_n\}$ are sequences such that $s_n\leq t_n$ then $$ \inf_n s_n\leq \inf_n t_n,\text{ and, }\sup_n s_n\leq \sup_n t_n.$$
- Proof of the Claim: (If you don't need it, you can scroll down). Let us show first that $\inf_n s_n\leq \inf_n t_n$. Suppose for a contradiction that $\alpha=\inf_n s_n > \inf_n t_n=\beta$, and consider $\epsilon=\alpha-\beta>0.$ By definition of inf, there is an element $t_n$ such that $$t_n\in (\beta,\beta+\epsilon)\Rightarrow\beta\leq t_n<\beta+\epsilon=\alpha\leq s_n\Rightarrow t_n<s_n,$$ contradicting the hypothesis that $s_n\leq t_n$. A similar proof can be done to show that $\sup_n s_n\leq \sup_n t_n$.
From this claim, we can prove that $\liminf_{n\to\infty}s_n\leq \liminf_{n\to\infty}t_n$ as follows:
Remember that $\liminf_{n\to\infty}s_n=\sup_{N\in\mathbb{N}}\inf_{n\geq N} s_n$, and similarly for $\liminf_{n\to\infty} t_n$. For $k\geq N$, consider the sequences given by $\sigma_k=\inf_{n\geq k}s_k$ and $\tau_k=\inf_{n\geq k}t_k$.
Since $s_n\leq t_n$ for every $n\geq N$, we have by the Claim that $\sigma_k\leq \tau_k$ for every $k\geq N$, and again by the claim,
$$\liminf_{n\to\infty} s_n=\sup_{k}\sigma_k\leq \sup_k \tau_k=\liminf_{n\to\infty}t_n.$$
A similar proof can be done from the definition of $\limsup$ changing a little bit the definition of the sequences $\sigma_k,\tau_k$, to finally show that $$\limsup_{n\to\infty}s_n=\inf_{k}\sigma_k\leq \inf_k\tau_k=\limsup_{n\to\infty} t_n.$$
Best Answer
For direct answer to your question let's consider $\mathbb{N}=\mathbb{N}_1\cup \mathbb{N}_2$ with $\mathbb{N}_1\cap \mathbb{N}_2=\emptyset$ and both infinite. If $f:\mathbb{N}_1\to \mathbb{R}$ and $f:\mathbb{N}_2\to \mathbb{R}$ are converged sequences, with limits $a_1$ and $a_2$ respectively, then any subsequence of $f:\mathbb{N}\to \mathbb{R}$ have only limit points $a_1$ and $a_2$.
This property can be extended to finite number of subsequences.
Addition.
To proof let's consider any subsequence $f_0:\mathbb{N}_0\to \mathbb{R}$ of sequence $f:\mathbb{N}\to \mathbb{R}$ i.e. some $\mathbb{N}_0\subset\mathbb{N}$, where $\mathbb{N}_0$ is infinite. Suppose $f_0$ have limit point $a_0$ different from $a_1$ and $a_2$. Then there exists a subsequence of $f_0$, which converges to $a_0$. But infinite number of members of this subsequence should be in $\mathbb{N}_1$ or $\mathbb{N}_2$, which is impossible, as they have their own different limit.