Prove that definitions of the limit superior are equivalent 
Let $(a_n)_{n\in\mathbb{N}}$ be a real sequence. And let $L^+$ be an extended real number (i.e. $L^+\in\mathbb{R}^*$). Then TFAE:
(1) $L^+$ = $\displaystyle\inf_{n\in\mathbb{N}}\sup_{k{\geq}n}(a_k)$ = $\displaystyle\lim_{n\to\infty}\sup_{k{\geq}n}(a_k)$
(2) $L^+$ = $\sup\{p\in\mathbb{R}^*:p$ is a cluster point of $(a_n)_{n\in\mathbb{N}}\}$
(3) $L^+$ is a (or the unique) number in $\mathbb{R}^*$ satisfying: (i) For all $\epsilon\gt 0$ there exists $N\in\mathbb{N}$ such that $a_n\lt L^++\epsilon$ for all $n\ge N$ (ii) For all $\epsilon\gt 0$ and $N\in\mathbb{N}$ there exists $n\ge N$ such that $a_n\gt L^+-\epsilon$ (or alternatively (i) For all $x\gt L^+$ there exists $N\in\mathbb{N}$ such that $a_n\lt x$ for all $n\ge N$ (ii) For all $x\lt L^+$ and $N\in\mathbb{N}$ there exists $n\ge N$ such that $a_n\gt x$)
The extended real number $L^+$ satisfying any of the above statement is called the limit superior of $(a_n)_{n\in\mathbb{N}}$

Question: To show these definitions are equivalent, I have no problem proving (1) $\Rightarrow$ (3) and (3) $\Rightarrow$ (2), but how to prove (2) $\Rightarrow$ (1)?
(If this is duplicate, please link to an answer really address this question.)
P.S. (1) is the definition according to Wikipedia and Terrence Tao's Analysis I, (2) is according to Rudin's Principles of Mathematical Analysis and Marsden's Elementary Classical Analysis, and (3) is according to Apostol's Mathematical Analysis
 A: Proof of (2) $\Rightarrow$ (1):
Denote $l$ = $\displaystyle\inf_{n\in\mathbb{N}}\sup_{k{\geq}n}(a_k)$ = $\displaystyle\lim_{n\to\infty}\sup_{k{\geq}n}(a_k)$
Suppose by way of contradiction that $L^+\gt l$
Then there exists $p\in\mathbb{R}^*$ such that (i) $p$ is a cluster point of $(a_n)_{n\in\mathbb{N}}$ and (ii) $p\gt l$
By (ii) we also have $\displaystyle p\gt\frac{1}{2}(p+l)\gt l$ = $\displaystyle\inf_{n\in\mathbb{N}}\sup_{k{\geq}n}(a_k)$
Hence $\displaystyle\sup_{k\ge K}a_k\lt\frac{1}{2}(p+l)$ for some $K\in\mathbb{N}$
Hence $\displaystyle a_k\lt\frac{1}{2}(p+l)$ for all $k\ge K$
Hence $\displaystyle\#\{k\in\mathbb{N}:|a_k-p|\lt\frac{1}{2}(p-l)\}\lt\infty$
Hence $p$ is not a cluster point of $(a_n)_{n\in\mathbb{N}}$. A contradicttion to (i)
Hence $L^+\le l$
Now given $\epsilon\gt 0$
There exists $N_{\epsilon}$ such that $\displaystyle l\le\sup_{k\ge N_{\epsilon}}a_k\lt l+\epsilon$
Since $\displaystyle(\sup_{k\ge j}a_k)_{j\in\mathbb{N}}$ is decreasing
We have $j\ge N_{\epsilon}$ implies (i) $\displaystyle l\le\sup_{k\ge j}a_k$ and (ii) $\displaystyle\sup_{k\ge j}a_k\lt l+\epsilon$
By (ii) we have $a_k\lt l+\epsilon$ for all $k\ge j$
By (i) we have $l\le a_{k_{j}}$ for some $k_j\ge j$
Hence $l\le a_{k_{j}}\lt l+\epsilon$ for some $k_j\ge j$ for all $j\ge N_{\epsilon}$
Hence $\#\{k\in\mathbb{N}:a_k\in(l-\epsilon, l+\epsilon)\}$ = $\infty$
Hence $l$ is a cluster point of $(a_n)_{n\in\mathbb{N}}$
Hence $L^+\ge l$
By the above argument we conclude that $L^+$ = $l$
