 Solution

A complete metric space is a metric space in which every Cauchy sequence converges to a point of that space.

If $x_n\to x$, choose $N$ so that $d(x_n,x)<\epsilon /2$ for all $n\ge N$. The triangle inequality gives $d(x_m,x_n)\le d(x_m,x)+d(x,x_n)<\epsilon$, proving the Cauchy sequence property.

Now let $A$ be complete in the induced metric. If $x\in \overline{A}$, choose $a_n\in A$ with $d(a_n,x)<1/n$. The sequence converges to $x$ in $X$, so is a Cauchy sequence. Completeness of $A$ supplies a limit $a\in A$. Uniqueness of limits in a metric space gives $a=x$, hence $\overline{A}\subseteq A$. Therefore a complete metric subspace is closed.

Completeness does not imply sequential compactness. In the complete space $\mathbb{R}$, the sequence $x_n=n$ has no convergent subsequence: a subsequence has strictly increasing indices $n_k\ge k$, so $x_{n_k}\to +\infty$ and is unbounded, whereas every convergent real sequence is bounded.