Looking for example of topological space , not co-countable topology , such that closure does not imply sequential closure I am looking for example of a topological space $X$ , whose topology is not the co-countable topology , such that there is some $A \subseteq X$ such that there exist some $a \in \bar A$ but there is no sequence in $A$ converging to $a$ .  
 A: Here’s an example different from the ones given in the answers to your earlier question.
Let $K=\{0\}\cup\left\{\frac1n:n\in\Bbb Z^+\right\}$ with the topology that it inherits from $\Bbb R$. For $k\in\Bbb Z^+$ let $$N_k=\{0\}\cup\left\{\frac1n:n\ge k\right\}\;;$$ the sets $N_k$ for $k\in\Bbb Z^+$ are a local base at $0$.
Let
$$Y=\{\langle 0,0\rangle\}\cup(\Bbb Z^+\times\Bbb Z^+)\;;$$
every point of $Y\setminus\{\langle 0,0\rangle\}$ is isolated in $Y$, and for each function $\varphi:\Bbb Z^+\to\Bbb Z^+$ we define a basic open nbhd $B(\varphi)$ of $\langle 0,0\rangle$ by
$$B(\varphi)=\{\langle 0,0\rangle\}\cup\{\langle m,n\rangle\in\Bbb Z^+\times\Bbb Z^+:n\ge\varphi(m)\}\;.$$
Now let $X=Y\times K$, and let $p=\big\langle\langle 0,0\rangle,0\big\rangle$. For each function $\varphi:\Bbb Z^+\to\Bbb Z^+$ let
$$A_\varphi=\left\{\left\langle\langle n,\varphi(n)\rangle,\frac1n\right\rangle:n\in\Bbb Z^+\right\}\;,$$
and let $A$ be the union of all the sets $A_\varphi$.
First, $p\in\operatorname{cl}_XA$. To see this, let $B(\varphi)\times N_k$ be any basic open nbhd of $p$ in the product $X$; then
$$\left\langle\langle k,\varphi(k)\rangle,\frac1k\right\rangle\in\big(B(\varphi)\times N_k\big)\cap A_\varphi\subseteq\big(B(\varphi)\times N_k\big)\cap A\;.$$
Next, no sequence in $A$ converges to $p$. To see this, let $\sigma=\langle x_k:k\in\Bbb Z^+\rangle$ be any sequence in $A$. For each $k\in\Bbb Z^+$ there are a function $\varphi_k:\Bbb Z^+\to\Bbb Z^+$ and an $n_k\in\Bbb Z^+$ such that $$x_k=\left\langle\langle n_k,\varphi_k(n_k)\rangle,\frac1{n_k}\right\rangle\;.$$ 
For each $m\in\Bbb Z^+$ let $S_m=\{k\in\Bbb Z^+:n_k=m\}$. There are two cases to be considered.


*

*There is an $m\in\Bbb Z^+$ such that $S_m$ is infinite. (In other words, infinitely many of the pairs $\langle n_k,\varphi_{n_k}(n_k)\rangle$ have the same first coordinate.) Then $Y\times N_{m+1}$ is a nbhd of $p$, and $x_k\notin Y\times N_{m+1}$ whenever $k\in S_m$, so $\sigma$ does not converge to $p$.

*All of the sets $S_m$ are finite. In this case define $\varphi:\Bbb Z^+\to\Bbb Z^+$ by $$\varphi(m)=\begin{cases}1,&\text{if }S_m=\varnothing\\
1+\max\{\varphi_k(n_k):k\in S_m\},&\text{otherwise}\;.\end{cases}$$ Then $B(\varphi)\times K$ is a nbhd of $p$, and $x_k\notin B(\varphi)\times K$ for each $k\in\Bbb Z^+$, so again $\sigma$ does not converge to $p$.
We’re not quite done, because $A$ need not be sequentially closed. However, if $A^+$ is the sequential closure of $A$, then $A^+$ is sequentially closed, and of course $p\notin A^+$. Moreover, $A\subseteq A^+$, so $p\in\operatorname{cl}_XA^+$, and $A^+$ is therefore not closed.
