Showing that $\displaystyle\limsup_{n\to\infty}x_n=\sup\{\text{cluster points of $\{x_n\}_{n=1}^\infty$}\}$

Let $\{x_n\}_{n=1}^\infty$ be a sequence in $\mathbb R$. I used to follow the definition that $$\limsup_{n\to\infty}x_n=\lim_{n\to\infty}\sup_{k\geq n}x_k$$ with limits understood in the sense of the extended real number system, but recently another definition has come into my sight:

If the sequence is not bounded above, we define $\limsup_{n\to\infty}x_n=\infty$. Otherwise, we define $$\limsup_{n\to\infty}x_n=\begin{cases} \sup(c\ell)&,c\ell\neq\emptyset\\ -\infty&,c\ell=\emptyset, \end{cases}$$ where $c\ell$ is the set of all cluster points of $\{x_n\}_{n=1}^\infty$. A cluster point of $\{x_n\}_{n=1}^\infty$ is a point in $\mathbb R$ whose open balls all contain infinitely many terms (some of them may repeat) of $\{x_n\}_{n=1}^\infty$.

I would like to show that this definition implies the old one:

If the sequence is not bounded above, then $\forall M\in\mathbb R$, $\exists N\in\mathbb N$ s.t. $x_N>M$. In this case, $\sup_{k\geq n}x_k=\infty$ for each $n\in\mathbb N$, implying $\{\sup_{k\geq n}x_k\}_{n=1}^\infty\searrow\infty$. Suppose instead that the sequence is bounded above. When the sequence has no cluster points, one can prove that $\lim_{n\to\infty}x_n=-\infty$, which in turn gives us $$\lim_{n\to\infty}\sup_{k\geq n}x_k=-\infty.$$ The proof has been doing well so far, but I have no idea how to show that if $c\ell\neq\emptyset$, then $$\lim_{n\to\infty}\sup_{k\geq n}x_k=\sup(c\ell).$$ The supremum of the cluster points? What is it really? Thank you.

Denote $E$ as the set of cluster points and $\beta=\sup(E)$. Denote $\alpha_n:=\sup_{k\geq n}x_k$, and assume this converges to some $\alpha$ from above. By definition, there exists $n_1$ such that $\alpha_1\geq x_{n_1}>\alpha_1-1/2.$ Given $i=1,2,\dots,k-1$, there exists $n_k>n_{k-1}$ such that $\alpha_k\geq x_{n_k}>\alpha_k-2^{-k}$. Since $\alpha_k \to \alpha,$ there exists $k_0$ such that $|\alpha_k-\alpha|<2^{-k}$ for all $k\geq k_0$.

Hence, $$ |x_{n_k}-\alpha|\leq |x_{n_k}-\alpha_k|+|\alpha_k-\alpha|\leq 2^{1-k} \to 0. $$ In order words, there exists a subsequence that converges to $\alpha.$ Hence, $\alpha$ is a cluster point so $\beta\geq \alpha$.

Since $E$ is closed, we must have $\beta \in E$ so there exists a subsequence $\{m_k\}$ with $x_{m_k}\to \beta$. But then for all $k$, we have $x_{m_k}\leq \alpha_{m_k}$. Hence, in the limit, we have $\beta \leq \alpha$.