Convergence implies lim sup = lim inf Could someone please explain to me how the following can be proven? I get the intution but don't know how to write it rigorously.

Thank you.
 A: Both $limsup$ and $liminf$ are limit points of a sequence; the largest and smallest limit points  respectively of a sequence, just as $lim$ is a limit point . Assuming your space is Hausdorff, or something else that guarantees that the limit is unique, then there is only one limit point, so we must have $liminf=limsup=lim$ 
A: The limit sup and inf of a sequence both are equal to the limit of some subsequence of that sequence. (Definition)
If $\lim_{n\to\infty} x_n=a$, then this limit is unique, that is, $\lim_{k\to\infty} x_{n_k}=a$, for all subsequences ${x_{n_k}}$. (Rudin thm. 3.2b)

proof of the theorem:
Suppose $p\in X$, $p'\in X$, and that {p_n} converges to $p$ and $p'$. And, suppose $\varepsilon\in\mathbb{R}^+$.
Then since the sequences converge, we can choose an $N$ with $n>N$ implying $d(p_n,p)<\varepsilon/2$ and an $N'$ with $n>N'$ implying $d(p_n,p')<\varepsilon/2$.
Then: $$d(p,p')\le d(p,p_n)+d(p_n,p')< \varepsilon$$
(where the first inequality holds from the triangle inequality.) Since this is true for every positive real number $\varepsilon$, we can conclude that $d(p,p')=0$, so that by the definition of a metric, $p=p'$.
