The limit points of an open interval (open set) I am not sure I understand why for the open set $(a,b)$, the limit points are $[a,b]$. Why are $a,b$ now included as limit points? Is this because we can somehow find a sequence in the open interval converging to these points?
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 $\epsilon &gt; 0$ then by construction we have

$$|x_n - a| &lt; \epsilon_n = \frac{d}{n}$$

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

$$|x_n - a| &lt; \epsilon_n = \frac{d}{n} &lt; \epsilon$$

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.
Suppose an open interval $(c, d)$ such that it contains only $a$, we can verify it is open in Euclidean topology (Topology Without Tears, definition 2.1.1), $(c, d)$ has no intersection with $(a, b)$, so, $a$ is not a limit point of $(a, b)$. Maybe I am wrong.