How does one prove that $1$ is a limit point using the definition for $p \in (0,1)$ and some $r>0$, $B_r(x)= \{q \in X: d(x,y) <r\}$? Let's suppose that we're given an open set $(0,1)$. How does one prove that $1$ is a limit point using the definition, for $p \in (0,1)$ and some $r&gt;0$, $B_r(x)= \{q \in X: d(p,q) &lt;r\}$?
Let $B_r(1)$ be a neighborhood of $1$. Assume without loss of generality that $r&lt;1$ (if $r\geqslant 1$ then just take the point, say $\frac12$). Then $|1 - (1-r/2)| = r/2&lt;r$, so $1-r/2\in B_r(1)$. Moreover, $1-r/2&gt;0$, so $1-r/2\in (0,1)$. Therefore $B_r(1)$ contains a point of $(0,1)$ other than $1$, so that $1$ is a limit point of $(0,1)$.
Depends on your definition of a limit point, so I am assuming that you are using the following definition: "A limit point $x$ of a set $S$ is a point such that for every $r&gt;0$, $B_r(x)\bigcap S\neq\emptyset$ and $B_r(x)\bigcap S\neq \{x\}$ ".

Edit: I forgot to say $B_r(x)\bigcap S\neq \{x\}$, but given you are working with real numbers it is impossible anyways.

Now consider $B_r(1) = (1-r,1+r)$. Is it possible that $B_r(1) \bigcap (0,1) = \emptyset$?