Question: If $x$ is a limit point of $A\subseteq \mathbb{R}$, then there exist a sequence in $A$ converging to $x$.
My efforts:
Take $\varepsilon = \frac1n$, $n\in \mathbb{N}$. As $x$ is a limit point, for every $n\in \mathbb{N}$ there exist an $x_n \in A-\{x\}$ such that $|x_n-x|<\varepsilon$. In other words every $\epsilon$-neighborhood of $x$ contains a point of $A$ other than $x$.
So what do I have now?
I have a sequence $\{x_{n}\}$ in $A$ and I want to show that limit of this sequence is $x$
Definition of Limit: Let $\{x_n\}$ be an sequence in $\mathbb{R}$, we say the sequence converges to $x$ if for every $\epsilon > 0$ there exist $N\in \mathbb{N}$ such that $|x_n-x|<\epsilon$ for every $N\geq \mathbb{N}$.
I am not able to use it to complete the last part of my proof. Intuitively I can see that $x_{n}$ converges to $x$.
It's not a good idea to state that $\varepsilon=\frac1n$ for every $n\in\mathbb N$. Is $\varepsilon$ a fixed number or not?
But you were almost there. For each $n\in\mathbb N$ take $x_n\in A$ such that $|x-x_n|<\frac1n$. I will prove that $\lim_{n\to\infty}x_n=x$. If $\varepsilon>0$, just take $N\in\mathbb N$ such that $\frac1N\leqslant\varepsilon$. Then$$n\geqslant N\implies|x-x_n|<\frac1n\leqslant\frac1N\leqslant\varepsilon.$$