Prove the set [a, b] is perfect

Question

How can I prove (by definition) that, if $a, b \in \mathbb{R}$ and $a<b$, then $[a, b]$ is equal to the set of accumulation (limit) points?

Let $(E, d)$ a metric space and $S \subseteq E$. $x \in E$ is a limit point if $(B_\varepsilon(x)-\lbrace x \rbrace ) \cap S \neq \emptyset$ for all $\varepsilon >0$

Answer 1

If $x \in [a,b]$, the $x \in B_\varepsilon(x) \cap [a,b]$ for all $\varepsilon >0$. If $x \notin [a,b]$ take $\varepsilon = \frac{1}{2}\min\{d(a,x),d(b,x)\}$, and notice that $B_\varepsilon(x) \cap [a,b] = \varnothing$, so $x$ is not a limit point of the set.

Answer 2

Attempt:

$U:= ${$x| x <a$}; $V:=${$x|x >b$};

$U,V$ are open sets,

1) $U \cup V$ is open.

$[a,b]=(U \cup V)^c$ is closed ($c$ for complement in $\mathbb{R}$).

2) Let $x_0 \not \in [a,b]$, then

$x_0 \in [a,b]^c= U \cup V$, which is an open set.

Hence

$B_\epsilon (x_0) \subset U \cup V$ for an $ \epsilon >0$.

$x_0$ is not a limit point of $[a,b]$.