= Solution
{c}
The answers are
$$
\boxed{\text{yes, no, yes, yes}.}
$$
Write $M_n(f,g)=\sup_{0\leq t\leq n}|f(t)-g(t)|$. The given <metric> describes <locally uniform convergence>: $d(f_j,f)\to0$ if and only if $M_n(f_j,f)\to0$ for every fixed $n$. Indeed,
$$
2^{-n}\min\{M_n(f_j,f),1\}\leq d(f_j,f),
$$
so convergence in $d$ forces $M_n\to0$. Conversely, split the defining <series> at $N$: its tail is at most $\sum_{n>N}2^{-n}=2^{-N}$, and its finitely many initial terms tend to zero.
If $(f_j)$ is a <Cauchy sequence> in $d$, the same single-term estimate makes it a <Cauchy sequence> in the <supremum norm> on every $[0,n]$. The <completeness of the continuous-map space in the uniform metric> gives a <continuous function> $F_n$ on $[0,n]$ with $f_j\to F_n$ uniformly there. Uniqueness of the <limit of a sequence> gives $F_{n+1}|_{[0,n]}=F_n$, so these <functions> define a single <continuous function> $f$ on $[0,\infty)$. The preceding finite-head and tail argument proves $d(f_j,f)\to0$. Thus the space is <complete>.
The constant <functions> $f_j(t)=j$ have $d(f_j,f_k)=1$ whenever $j\ne k$. They have no <convergent subsequence>, so the space is not <sequentially compact>.
For fixed $f,g$, set $\gamma(s)=(1-s)f+sg$. On $[0,n]$,
$$
M_n(\gamma(s),\gamma(s'))=|s-s'|M_n(f,g)\longrightarrow0
$$
as $s'\to s$. The equivalence with <locally uniform convergence> proves that $\gamma$ is a <continuous path> for $d$. Hence the space is <path-connected> and <connected>.
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