$A\subseteq \mathbb R^2$ and $X=\mathbb R^2\backslash A$ are both equipped with subspace topology inherited from $\mathbb R^2.$ If $A$ is countable and dense, does that imply that necessarily $X$ is totally disconnected $?$
If I take $A=\mathbb Q\times \mathbb Q$ then $A$ is countable dense and the complement $X=\mathbb Q^c\times \mathbb Q^c$ is totally disconnected. But is this always true I mean if I take any arbitrary $A$ from $\mathbb R^2$ qwill this hold as a result $?$
The complement of $A$ is not totally disconnected - your equation $$(C\times C)^c=(C^c)\times (C^c)$$ is incorrect (e.g. the point $(\pi, 7)$ is in $A^c$ but not in $\mathbb{Q}^c\times \mathbb{Q}^c$). Indeed, the complement of $A$ is connected, as is any co-countable subset of $\mathbb{R}^2$.