Here is one approach (another being this).

Let $ (x_n) $ be a bounded sequence of reals. Thanks to Bolzano-Weierstrass, $ A := \lbrace \text{ accumulation points of }(x_n)   \rbrace $ is non-empty. $ A $ is also bounded above, so $ a := \sup(A) $ exists, which we call $ \color{goldenrod}{\limsup x_n} $.

$ a \in A $, i.e. $ a $ itself is an accumulation point.

Let $ \epsilon > 0 $. There is an accumulation point $ p $ in $ (a-\epsilon, a ] $. Now $ x_n \in (p-\epsilon, p+\epsilon) $ for infinitely many $ n $. So $ (a-2\epsilon, a+2\epsilon) $ $ \supseteq (p-\epsilon, p+\epsilon) $ contains $ x_n $ for infinitely many $ n $.

Let $ \epsilon > 0 $.
$ x_n > a+\epsilon $ for only finitely many $ n $ : if there were infinitely many, by BW there will be an accumulation point $ > a $, absurd.
$ x_n > a-\epsilon $ for infinitely many $ n $ : there are infinitely many $ n $ for which $ x_n \in (a-\epsilon, a+\epsilon) $.

Interestingly the above para characterises $ a $.
Say $ l \in \mathbb{R} $ is such that for every $ \epsilon > 0 $, $ x_n > l+\epsilon $ for only finitely many $ n $ and $ x_n > l -\epsilon $ for infinitely many $ n $. Then $ l = a $.

If $ l > a $, as $ x_n > l - \frac{l-a}{2} $ for infinitely many $ n $, we get $ x_n > a + \frac{l-a}{2} $ for infinitely many $ n $, absurd.
If $ l < a $, as $ x_n > a - \frac{a-l}{2} $ for infinitely many $ n $, we get $ x_n > l + \frac{a-l}{2} $ for infinitely many $ n $, absurd.
So $ l = a $.