Integrating the two-views of lim sup and lim inf

Preliminaries

Let $(x_n)$ be a bounded real number sequence and $(x_n )_{n≥k}$be a subsequence of $x_n$ which only takes the values of the sequence starting from the k−th term.

Let {$x_n$} and {$x_n$ }$_{n≥k}$ denote a subset of R that contains all values of the corresponding sequences $(x_n )$ and $(x_n )_{n≥k}.$

Initial Considerations
Define two sequences ($\alpha_k$) and ($\beta_k$) out of the original ($x_n$) sequence :
($α_k$)= sup{$x_n$}$_{n≥k}$
($\beta_k$)= inf{$x_n$}$_{n≥k}$

Call a number $c \in R$ a subsequential limit of the real number sequence ($x_n$) iff there exists a subsequence of ($x_n$) that converges to c.

Define C $\in R$ as the set of all subsequential limits of ($x_n$).

Definition of Lim sup and Lim inf

1 - Define lim sup $x_n$ as the supremum of C and define lim inf $x_n$ as the infimum of C.

2 - Define lim sup $x_n$ as the infimum of {$\alpha_k$} and lim inf $x_n$ as the supremum of {$\beta_k$}.

My doubt

Why the supremum of C is the same as the infimum of {$\alpha_k$} and the infimum of C is the same as the supremum of {$\beta_k$} ?

In another words, what's special about the construction of the ($\alpha_k$) and ($\beta_k$) sequences that make their limits ( their infimums/supremums because they are bounded monotone decreasing/increasing, respectively ) to be largest/smallest possible subsequential limits of the original sequence sequence ($x_n$) ?

Can i understand the equivalence between these two definitions ( preferably both intuitively and rigorously ) without touching on anything related to topology ?

The reason is that my real analysis teacher ( who has not taught us anything about topology yet ) is teaching us sequences using the second definition of lim sup and lim inf and many of the theorems we are going through are better understood ( in my view ) if one understands both definitions of lim sup and lim inf ( and obviously, their equivalence ).

I can show you a proof that if you define $\limsup x_n:=\lim_{n\to \infty} \sup\{x_k\}_{k\ge n}$, then $\limsup x_n=C$, where $C$ is the supremum of all subsequential limits. First, see this question for proofs that $\limsup x_n\ge C$.
To show that $\limsup x_n\le C$, it suffices to show there is a subsequence $(x_{n_m})$ that converges to $\limsup x_n$. This subsequence can be constructed inductively. Put $x_{n_1}=x_1$. Let $\alpha_k:=\sup \{x_k\}_{k\ge n}$. Now, suppose $x_{n_1},\ldots x_{n_{m-1}}$ have been defined. There must be some $k_0\ge n_{m-1}$ such that $\alpha_{n_{m-1}}-\frac1m\le x_{k_0}\le \alpha_{n_{m-1}}$. Put $n_m:=k_0$. This completes the construction of $(x_{n_m})$. Now, it follows from the construction that $|x_{n_m}-\alpha_{n_{m-1}}|<\frac1m$ for all $m\in \mathbb{N}$. Since $\lim_{m\to \infty}\alpha_{n_{m-1}}=\limsup x_n$, it follows that $\lim_{n\to \infty} x_{n_m}=\limsup x_n$.
Well, I find the relation (limsup by def. 1) $\le$ (limsup by def. 2) pretty intuitive - if there was a $c \in C$ such that inf {$\alpha_k$} $<$ c, that would mean that there is a k such that sup{$x_n$}$_{n \ge k} < c$, but since a subsequence of $x_n$ gets arbitrarily close to $c$, that must be a contradiction.
To see that (limsup by def. 2) = $\alpha$ $\in C$ you just have to consider that for any $k$ there are elements of {$x_n$}$_{n \ge k}$ abritrarily close to $\alpha$, and so you can "pick" a subsequence of $x_n$ that will tend to $\alpha$.