Question: $A= \{x∈\mathbb{R}: \tan{x}\text{ is not differentiable}\}$ is complete?
My attempt: clearly $A=\{(2n+1)\frac{π}{2}: n∈\mathbb{Z}\}$.
I can see each entry of $A$ is irrational, and we know between two irrational there exists rational number(hence infinity of rationals!) which is limit point of $A$ but does not belongs to $A$. Hence $A$ is not closed! But, $A$ subset of complete metric space $\mathbb{R}$, so is that imply $A$ is Not complete?
Your argument about there being a rational between any two (distinct) irrationals doesn't really make a lot of sense, in that this fact doesn't show that such rational numbers are limit points of $A$ (they're not).
You know that a subspace of a complete metric space is also complete if and only if it is closed. Indeed, $A$ is closed; its complement is a union of open sets $\left((2n+1)\frac{\pi}{2},(2n+3)\frac{\pi}{2}\right)$ and hence open.
Another way of thinking about it is that $A$ is uniformly discrete, which is to say that there is a fixed $\delta>0$ such that the distance between any two distinct elements of $A$ is $\geq \delta$. In general, if a metric space is uniformly discrete, it will be complete, as the given metric will be strongly equivalent to the standard discrete metric, and discrete metric spaces are complete (Cauchy sequences in such space must be eventually constant so convergent).