 Solution

Let $a^{(m)}=(1,1/2,\dots ,1/m,0,0,\dots )$. Each has finite support, and in the supremum norm metric it converges to $a=(1,1/2,1/3,\dots )$, because $d(a^{(m)},a)=1/(m+1)$.

The limit is bounded but does not have finite support. Thus the finite-support subspace is not closed in $X$, and by part (a)(iii) it is not complete. Equivalently, the displayed sequence is Cauchy in that subspace but has no limit there.