The limit points of an open interval (open set)

Indeed we can find sequences that converge to those points.

Let $d = b-a$ and $\epsilon_n = \frac{d}{n}$. For each $n \in \mathbb{N}$ choose a point $x_n \in (a, a+\epsilon_n)$. Now we claim that that the sequence $\{x_n\}$ converges to $a$.

Let $\epsilon > 0$ then by construction we have

$$|x_n - a| < \epsilon_n = \frac{d}{n}$$

Thus we can let $N\in \mathbb{N}$ with $N > \frac{d}{\epsilon}$ so that for all $n \ge N$ we have

$$|x_n - a| < \epsilon_n = \frac{d}{n} < \epsilon$$

which means that $\lim x_n = a$. Thus $a$ is a limit point of $(a,b)$ and by similar approach (which I will leave to you), it is possible to show that $b$ is also a limit point.

$\boxtimes$

So what we have really done here is look at smaller and smaller intervals contained in $(a,b)$ that are close to $a$ and pick points from each of these intervals. These points form our sequence and the limit of this sequence is $a$. If you need help on constructing a sequence that converges to $b$, think about what intervals you would need to pick points from.

$a$ is a limit point

Sequence $x_n=a+\frac{1}{n} \in (a,b)\forall n \in \mathbb N(\because$ consequence of archemedean property). We know that $x_n \to a$ as $n \to \infty$. Hence $a$ is the limit point of $(a,b)$. ($\because$ $x\in \mathbb{R}$ is a limit point of $A$ iff there is a sequence of points $\{x_{n}\}\subset A$ which are different from $x$ and converges to $x$). Similar argument for $b$.

So according to your definition of limit points, it sounds like the points in $\overline{A}$, if so, yes, because the points $x_{n}=a+\dfrac{b-a}{2n}$, $n=1,2,...$ belong to $(a,b)$ and are such that $x_{n}\rightarrow a$.