= Solution
{c}
The space of <polynomials> is not <complete> for the <supremum norm> <metric> $d$. Consider
$$
p_n(t)=\sum_{k=0}^n(t/2)^k.
$$
The <geometric series> has sum $F(t)=1/(1-t/2)$, and for $0\leq t\leq1$ its tail satisfies
$$
|F(t)-p_n(t)|=\frac{(t/2)^{n+1}}{1-t/2}\leq2^{-n}.
$$
Thus $(p_n)$ converges uniformly to $F$ and is a <Cauchy sequence> for $d$.
There is no <polynomial> $p$ agreeing with $F$ on $[0,1]$: otherwise the <polynomial> $(1-t/2)p(t)-1$ would vanish at infinitely many points and therefore be the zero <polynomial>. Evaluating this identity at $t=2$ gives $-1=0$. By uniqueness of the <uniform limit>, $(p_n)$ has no limit in $X$. This proves the claimed failure of <completeness>.
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