Let $(X ,\mathfrak T)$ be a topological space and suppose that $A$ is a subset of $X$. Then $Cl(A) = A\cup Bd(A)$. False! Let $(X ,\mathfrak T)$ be a topological space and suppose that $A$ is a subset of $X$. Then $Cl(A) = A\cup Bd(A)$.
I think this statement is false because the definition of closure does have the union of A and the limit points of A. However, I know that it is possible to have boundary points that are not limit points. 
If I am correct what is a basic example that models this?
 A: This is not a false claim. By (one) definition of boundary we have $\text{Bd}(A) = \overline{A}\setminus A^\circ$ where $A^\circ$ is the interior of $A$. Hence $$A \cup (\overline{A}\setminus A^\circ) = \overline{A}$$
We have equality from the union above since $A^\circ \subset A$. Hence $$A\cup \text{Bd}(A) = \text{Cl}(A)$$

Edit: As requested, here is how you could prove this result with sets, elements and double containment. 
Let $x \in \overline{A}$. It should be clear that either $x \in A^\circ$ or $x \notin A^\circ$. In the latter case, this means $x\in \overline{A}/A^\circ$ so clearly $x\in A\cup \overline{A}/A^\circ$. In the former case, we can use the fact that $A^\circ \subset A$ to see that $x \in A$ so $x \in A \cup \overline{A}/A^\circ$. Hence, $\overline{A} \subset A\cup \overline{A}/A^\circ$.
Now let $y \in  A\cup \overline{A}/A^\circ$ and suppose FTSOC that $y \notin \overline{A}$. Since $A,A^\circ \subset \overline{A}$ this means $y \notin A$ and $y\notin A^\circ$, so $x\notin A \cup \overline{A}$. It is obvious that $A\cup \overline{A} /A^\circ \subset A \cup \overline{A}$ so  $y \notin A\cup \overline{A} /A^\circ$, a contradiction. Thus $\overline{A} = A \cup \overline{A}/A^\circ$.
A: You might be getting confused with the following: given a topological space $A$, it is not necessarily the case that every point of the closure $\overline{A}$ is the topological limit of some sequence $a_1,a_2,a_3,\dots$ of points of $a$.  
Indeed, let $R$ be an uncountable set, and consider the space $X=\{f\colon R\to\{0,1\}\}$.  For a countable set $B\subset A$ and a function $g\colon A\to\{0,1\}$, we define
$$
U_{B,g}=\{f\in X\;\colon\; f(\alpha)=g(\alpha)\textrm{ for all }\alpha\in B\}
$$
The sets $U_{B,g}$ form the basis for a topology on $X$.  Now define a subset $Y$ of $X$ by
$$
Y=\{f\in X\;\colon\;f(\alpha)=0\textrm{ for all but countably many }\alpha\in A\}
$$
Now if $g_1,g_2,g_3,\dots$ is a sequence in $Y$ such that $g_n\to g\in X$, then $g\in Y$.  So $Y$ contains all its limit points.  On the other hand, $Y$ is not closed, since it is dense in $X$.  
