$A$ is compact iff $ A $ is bounded and, given $\epsilon > 0$, there exists $ n_0$ such that $ \sum_ {k=n}^\infty |x_k|\le\epsilon $ for all $n \geq n_0 $ and for all $ x\in A $.
To prove sufficient condition for compactness, I try to prove $A$ is totally bounded and closed. I am able to prove only the totally bounded condition. Now how do I prove $ A $ is closed?
Under these hypotheses, $A$ need not be closed (nor compact). Let $x$ be the sequence $(1,0,0,\dots)$ (i.e. $x_1 = 1$ and $x_k = 0$ for $k \ge 2$). Let $A= \{ cx : 0 < c < 1\}$. $A$ is bounded and satisfies your uniform integrability condition (with $n_0 = 2$), but it is not closed since it has 0 as a limit point but does not contain it.
However, you should be able to show that under these conditions, the closure of $A$ is compact, or in other words $A$ is precompact (or relatively compact).