Solution: Let us write three equivalent definitions of boundary of a set $A \subseteq X$. Let $\bar{A} $ denotes the closure of $A$ in $X$ and $\mathrm{int}(A) $ denotes the interior of $A$ in $X$.

1. Definition given in the problem.
2. $\partial A = \bar{A} \setminus \mathrm{int}(A) $.
3. $\partial A = \bar{A} \cap \overline{\left( X\setminus A \right)}$.

Consider the following subsets:

1. Take $A=\mathbb{Z} $. We claim that $\partial \mathbb{Z} =\mathbb{Z} $. Let $n\in \mathbb{Z} $. Take any neighborhood of $n$, say $U$ is an open neighborhood of $n$. We have \[ n\in U\cap \mathbb{Z}, \] so, $U$ intersects $\mathbb{Z} $. Now we need to show that it intersects $\mathbb{Z} ^c$. But this is clear as $U$ is an open subset of $\mathbb{R} $ containing $n$ and hence there exists $\varepsilon >0$ such that $(n-\varepsilon ,n+\varepsilon )\subseteq U$. Therefore, $U$ intersects $\mathbb{Z} ^c$ . Thus, $n\in \partial Z$ and hence, $\partial \mathbb{Z} \subseteq \mathbb{Z} $.
   
   On the other hand, if $x\in \partial \mathbb{Z} $, and $x\notin \mathbb{Z} $, then let $n$ be the integer such that $n \lt x \lt x+1$. Here $n=[x]$, where $[x]$ is the greatest integer function of $x$. Take \[ \varepsilon = \min \left\{ x-n, n+1-x \right\}. \]
   
   Then, the open set $U = (x-\varepsilon ,x+\varepsilon )$ does not intersect $\mathbb{Z} $ and hence we $x\notin \partial \mathbb{Z} $. Therefore, we got a contradiction. Hence, if $x\in \partial \mathbb{Z} $, then $x\in \mathbb{Z} $. This proves that $\partial \mathbb{Z} \subseteq \mathbb{Z} $ and hence, $\partial \mathbb{Z} = \mathbb{Z} $.
2. Take $A = \mathbb{Z} \cup (1,2)$. Let us prove that the boundary $A$ is $\mathbb{Z} $ with the second definition. Recall that \[ \overline{A \cup B} = \bar{A} \cup \bar{B} \] \begin{align*} \bar{A} & = \overline{\mathbb{Z} \cup (1,2)} \\ & = \overline{\mathbb{Z} } \cup \overline{(0,1)} \\ & = \mathbb{Z} \cup [1,2] \\ & = \mathbb{Z} \cup (1,2). \end{align*} We need to determine the interior of $\mathbb{Z} \cup (1,2)$. We claim that \[ \mathrm{int}\left( \mathbb{Z} \cup (1,2) \right) = (1,2). \] It is clear that $(1,2)\subseteq \mathrm{int}\left( \mathbb{Z} \cup (1,2) \right) $. For the reverse inequality, let $x\in \mathrm{int}\left( \mathbb{Z} \cup (1,2) \right) $. We need to show that $x\in (1,2)$. Suppose that $x\notin (1,2)$. Since we know that for any subset $A$ we have $\mathrm{int}(A) \subseteq A $. Therefore, $\mathrm{int}\left( \mathbb{Z} \cup (1,2) \right) \subseteq \mathbb{Z} \cup (1,2)$. Since $x\notin (1,2)$, so $x$ must be an integer. Thus any open neighborhood of $x$ does not contain in $A$ and hence $x$ can not be an interior point. Therefore, $x\in (1,2)$. Hence, \[ \mathrm{int}\left( \mathbb{Z} \cup (1,2) \right) = (1,2). \] Therefore, \begin{align*} \partial A & = \bar{A} \setminus \mathrm{int} (A) \\ & = \left( \mathbb{Z} \cup (1,2) \right) \setminus (1,2) \\ & = \mathbb{Z} . \end{align*}
3. Take $A= \mathbb{Z} \cup (2,3)$.