 Solution

The answers areDistinct natural numbers have $d(x,y)>1$. Taking $\epsilon =1$ in the Cauchy sequence condition forces every sufficiently late pair of terms to be equal. Every Cauchy sequence is therefore eventually constant and converges in $\mathbb{N}$, proving completeness.

The sequence $x_n=n$ has no convergent subsequence, since convergence would force its sufficiently late terms to lie within distance $1$ of its limit of a sequence, hence to equal that limit. Thus sequential compactness fails.

Every singleton is an open ball of radius $1$, so the metric topology is the discrete topology. For example, $\{1\}$ and $\mathbb{N}\setminus \{1\}$ form a separation into two nonempty open sets, so $X$ is not connected. It is consequently not path-connected either.