$\mathbb{Z}$ (in $\mathbb{R}$) has no limit points (sequential) Our definition of limit point is that $p$ is a limit point if there exists a non-constant sequence in $\mathbb{Z}$ that converges to $p$.
How can I prove that there don't exist any sequences in $\mathbb{Z}$ that converge?
 A: Use the definition of convergence: choose a small neighborhood around $p$ that doesn't contain any other integers, and show that any sequence of integers satisfying the convergence criteria is necessarily constant.
(Actually, precisely, it is eventually constant; for instance, the sequence $\{1,2,2,2,\dots,2,\dots\}$ is non-constant, and converges to $2$, but is not what you want to allow here)
A: Your definition of limit point is wrong. The sequence
$$
a_n=\begin{cases}
0 & \text{if $n=0$}\\
1 & \text{if $n>0$}
\end{cases}
$$
is not constant and clearly converges to $1$. But $1$ is not a limit point of $\{0,1\}$ in $\mathbb{R}$, because $\{0,1\}$ consists of two isolated points.
The correct definition is

$p$ is a limit point of $S\subseteq \mathbb{R}$ if there exists a non eventually constant sequence in $S$ converging to $p$.

A sequence $(a_n)$ is eventually constant if there is $N$ such that, for $n\ge N$, $a_n=a_N$.
Now, if $(a_n)$ is a sequence in $\mathbb{Z}$ that converges to $p$, it is eventually constant. Indeed, there is $N$ such that, for $n\ge N$, $|a_n-p|<\frac{1}{4}$. In particular, for $n\ge N$,
$$
|a_N-a_n|=|(a_N-p)+(p-a_n)|\le|a_N-p|+|p-a_n|<\frac{1}{4}+\frac{1}{4}=\frac{1}{2}
$$
and therefore $a_n=a_N$. So the sequence is eventually constant.
A: Hint
If a sequence in $\Bbb R$ converges, it is a Cauchy sequence, but if the terms of a sequence $\{a_n\}$ are different integers, then $|a_p-a_q|\ge 1$.
