Theorem 3.19 in Baby Rudin: The upper and lower limits of a majorised sequence cannot exceed those of the majorising one Here is Theorem 3.19 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition.

Let $\{s_n \}$ and $\{t_n \}$ be sequences of real numbers. If $s_n \leq t_n$ for $n \geq N$, where $N$ is fixed, then $$\lim_{n\to\infty} \inf s_n \leq \lim_{n\to\infty} \inf t_n,$$ $$\lim_{n\to\infty} \sup s_n \leq \lim_{n\to\infty} \sup t_n.$$

Rudin has not given a proof for this theorem, which he considers to be ''quite trivial".
I'll only give the proof for the limit superior, and I'll be using Theorem 3.17.
Now here's my proof:

If $\lim_{n \to \infty} \sup s_n > \lim_{n \to \infty} \sup t_n $, then let's choose a real number  $x$ such that $\lim_{n \to \infty} \sup s_n > x > \lim_{n \to \infty} \sup t_n $.
Then by part (b) of Theorem 3.17, there is an integer $N_1$ such that $n \geq N_1$ implies that $t_n < x$.
So, $n > \max \{ N, N_1 \}$ implies that $s_n \leq t_n < x$.
Now if we take a real number $\varepsilon$ such that $$0 < \varepsilon < \frac{\lim_{n\to\infty} \sup s_n \ - x}{3},$$ then the neighborhood $$\left( \lim_{n\to\infty} \sup s_n - \varepsilon, \ \lim_{n\to\infty} \sup s_n + \varepsilon \right)$$ of $\lim_{n\to\infty} \sup s_n$ contains at most a finite number of temrs of the sequence $\{s_n\}$, which shows no subsequence of $\{s_n \}$ can converge to $\lim_{n\to\infty} \sup s_n$ (or diverge to $+\infty$ if $\lim_{n\to\infty} \sup s_n= +\infty$), which is a contradiction to part (a) of Theorem 3.17.

Is this proof correct? If not, where does the problem lie? If this proof is correct, then is this the proof that Rudin would be requiring?
In response to the request of a Stack Exchange member, I'll include Rudin's definitions of these concepts.
Definition 1.8:

Suppose $S$ is an ordered set, $E \subset S$, and $E$ is bounded above. Suppose there exists an $\alpha \in S$ with the following properties:
(i) $\alpha$ is an upper bound of $E$.
(ii) If $\gamma < \alpha$, then $\gamma$ is not an upper bound of $E$.
Then $\alpha$ is called the least upper bound of $E$ [that there is at most one such $\alpha$ is clear from (ii)] or the supremum of $E$, and we write $$\alpha = \sup E.$$
The greatest lower bound, or infimum, of a set $E$ which is bounded below is defined in the same manner: The statement $$\alpha = \inf E$$ means that $\alpha$ is a lower bound of $E$ and that no $\beta$ with $\beta > \alpha$ is a lower bound of $E$.

De3finition 1.23:

The extended real number system consists of the real field $R$ and two symbols, $+\infty$ and $-\infty$. We preserve the original order in $R$, and define $$-\infty < x < +\infty$$ for every $x \in \mathbb{R}$.
It is then clear that $+\infty$ is an upper bound of every subset of the extended real number system, and that every nonempty subset has a least upper bound. If, for example, $E$ is a nonempty set of real numbers which is not bounded above in $R$, then $\sup E = +\infty$ in the extended real number system.
Exactly the same remarks apply to lower bounds.

Definition 3.5:

Given a sequence $\{ p_n \}$, consider a sequence $\{n_k \}$ of positive integers, such that $n_1 < n_2 < n_3 < \cdots$. Then the sequence $\{p_{n_k}\}$ is called a subsequence of $\{p_n\}$. If $\{p_{n_k}\}$ converges, its limit is called a subsequential limit of $\{p_n\}$.
It is clear that $\{p_n \}$ converges to $p$ if and only if every subsequence of $\{p_n \}$ converges to $p$.

Theorem 3.6:

(a) If $\{ p_n \}$ is a sequence in a compact metric space $X$, then some subsequence of $\{p_n \}$ converges to a point of $X$.
(b) Every bounded sequence in $\mathbb{R}^k$ contains a convergent subsequence.

Theorem 3.7:

The subsequential limits of a sequence $\{p_n\}$ in a metric space $X$ form a closed subset of $X$.

Definition 3.15:

Let $\{s_n \}$ be a sequence of real numbers with the following property: For every real $M$ there is an integer $N$ such that $n \geq N$ implies $s_n \geq M$. We then write $$s_n \rightarrow +\infty.$$ Similarly, if for every real $M$ there is an integer $N$ such that $n \geq N$ implies $s_n \leq M$, we write $$s_n \rightarrow -\infty.$$

Definition 3.16:

Let $\{s_n\}$ be a sequence of real numbers. Let $E$ be the set of numbers $x$ (in the extended real number system) such that $s_{n_k} \rightarrow x$ for some subsequence $\{s_{n_k}\}$. This set $E$ contains all subsequential limits as defined in Definition 3.5, plus possibly the numbers $+\infty$, $-\infty$.
We now recall Definitions 1.8 and 1.23 and put $$s^* = \sup E,$$ $$s_* = \inf E.$$ The numbers $s^*$, $s_*$ are called the upper and lower limits of $\{s_n \}$; we use the notation $$\lim_{n\to\infty} \sup s_n = s^*, \ \ \ \lim_{n\to\infty} \inf s_n = s_*.$$

Theorem 3.17:

Let $\{ s_n \}$ be a sequence of real numbers. Let $E$ and $s^*$ have the same meaning as in Definition 3.16. Then $s^*$ has the following two properties:
(a) $s^* \in E$.
(b) If $x > s^*$, then there is an integer $N$ such that $n \geq N$ implies $s_n < x$.
Moreover, $s^*$ is the only number with the properties (a) and (b).

I've tried to reproduced from Rudin's text some of the more  relevant theorems and definitions.
 A: 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 $\{x_n\}$, $$-\liminf_{n\to\infty} x_n=\limsup_{n\to\infty} (-x_n).$$
Then you have the other one by the one we proved above.
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 $$\limsup_{n\to\infty}s_n=\inf_{k}\sigma_k\leq \inf_k\tau_k=\limsup_{n\to\infty} t_n.$$
A: If $s_n\leq t_n$ for each $n\geq N$ then evidently for each $n\geq N$:  $$\sup_{k\geq n}s_k\leq\sup_{k\geq n}t_k$$
If $(a_n)$ and $(b_n)$ are both non-increasing sequences in $(-\infty,+\infty]$ then their limits exist in $[-\infty,+\infty]$. If moreover $a_n\leq b_n$ for each $n\geq N$ then evidently: $$\lim_{n\to\infty}a_n\leq\lim_{n\to\infty}b_n$$
Now take $a_n=\sup_{k\geq n}s_k$ and $b_n=\sup_{k\geq n}t_k$ to end up with: $$\limsup s_n=\lim_{n\to\infty}\sup_{k\geq n}s_k\leq\lim_{n\to\infty}\sup_{k\geq n}t_k=\limsup t_n$$

edit:
Proof based on def. 3.16
First of all: every sequence $\left(a_{n}\right)_{n}$ in $\mathbb{R}$
will have a subsequence $\left(a_{n_{k}}\right)$ such that $a_{n_{k}}\to z$
for some $z\in\left[-\infty,+\infty\right]$. This is actually the statement that the set $E$ mentioned under def. 3.16 is not empty.
If you would like to see this proved as well then please let me know.
Assume that $t^{*}<s^{*}$. Then by definition $\left(s_{n}\right)_{n}$ must have
a subsequence $\left(s_{n_{k}}\right)_{k}$ with $s_{n_{k}}\to x$
for some $x>t^{*}$. Now let $\left(t_{n_{k_{i}}}\right)$ be a subsequence
of $\left(t_{n_{k}}\right)$ with $t_{n_{k_{i}}}\to y$. On base of
$t_{n_{k_{i}}}\geq s_{n_{k_{i}}}$ for each $i$ we conclude that
$y\geq x>t^{*}$. 
But $\left(t_{n_{k_{i}}}\right)$ is also a subsequence
of $\left(t_{n}\right)$, so $y\leq t^{*}$ and a contradiction is
found.
A: My proof is not elegant, but this answer is most natural for me and a memo for me.  

Let $\{s_n \}$ and $\{t_n \}$ be sequences of real numbers. If $s_n \leq t_n$ for $n \geq N$, where $N$ is fixed, then $$\lim_{n\to\infty} \inf s_n \leq \lim_{n\to\infty} \inf t_n,$$ $$\lim_{n\to\infty} \sup s_n \leq \lim_{n\to\infty} \sup t_n.$$ 

(1)
If $\limsup_{n\to\infty}s_n = +\infty$, then, there exists a subsequence $\{s_{n_k}\}$ such that $s_{n_k} \to +\infty$.
Then, $s_{n_k} \leq t_{n_k}$ for all sufficiently large $k$.
So, $t_{n_k} \to +\infty$.
So, $\limsup_{n\to\infty}s_n = +\infty = \limsup_{n\to\infty}t_n$.  
(2)
If  $\limsup_{n\to\infty}s_n = -\infty$, then $\limsup_{n\to\infty}s_n = -\infty \leq \limsup_{n\to\infty}t_n$.  
(3)
We assume that $\limsup_{n\to\infty}s_n = s^* \in \mathbb{R}$.
Then, there exists a subsequence  $\{s_{n_k}\}$ such that $s_{n_k} \to s^*$.
(3-1)
If $\limsup_{n\to\infty}t_n = +\infty$, then $\limsup_{n\to\infty}s_n < +\infty = \limsup_{n\to\infty}t_n$.
(3-2)
If $\limsup_{n\to\infty}t_n = -\infty$, then $\lim_{n \to \infty} t_n = -\infty$ by Theorem 3.17(b).
So, $s_n \to -\infty$.
So, $\limsup_{n\to\infty}s_n = -\infty = \limsup_{n\to\infty}t_n$.
(3-3)
We assume that $\limsup_{n\to\infty}t_n = t^* \in \mathbb{R}$.
Let $\epsilon$ be an arbitrary positive real number.
By Theorem 3.17(b), for all sufficiently large $n$, $t_n < t^* + \epsilon$.
So, for all sufficiently large $n$, $s_n \leq t_n < t^* + \epsilon$.
So, for all sufficiently large $k$, $s_{n_k} < t^* + \epsilon$.
$\{s_{n_k}\}$ is a sequence such that $s_{n_k} \to s^*$.
So, $s^* \leq t^* + \epsilon$.
$\epsilon$ is an arbitrary positive real number.
So, $\limsup_{n\to\infty}s_n = s^* \leq t^* = \limsup_{n\to\infty}t_n$.  
