Prove/Disprove the closure of open ball is complete/ totally bounded/ compact Consider the metric spaces of all continuous functions on [0,1] mapping into R. The metric is given by $$d(f,g)=\int_0^1|f(t)-g(t)|dt.$$
Prove of disprove the following statements directly.
(i) $B\overline(0,1)$ is complete.
(ii) $B\overline(0,1)$ is totally bounded.
(iii) $B\overline (0,1)$ is compact.
(iv) every sequence in $B\overline (0,1)$ has a subsequence that converges in $B\overline (0,1)$.

I am a freshman of this real analysis course and really confused. Thanks to anyone who is coming for help!
 A: Hints
(i) Prove that functions
$$
f_n(t)=\begin{cases}
0&t\in\left(0,\frac{1}{2} -\frac{1}{2n}\right)\\
\frac{1}{2}+n\left(t-\frac{1}{2}\right) &t\in\left[\frac{1}{2}-\frac{1}{2n},\frac{1}{2}+\frac{1}{2n}\right]\\
1 &t\in\left(\frac{1}{2}+\frac{1}{2n},1\right)
\end{cases}
$$
forms non-convergent Cauchy sequence.
(iv) Consider "peak" functions
$$
g_n(t)=\begin{cases}
4^{n+1}\left(t-\frac{1}{2^{n+1}}\right)&t\in\left[\frac{1}{2^{n+1}},\frac{3}{2^{n+1}}\right]\\
-4^{n+1}\left(t-\frac{1}{2^{n}}\right) &t\in\left[\frac{3}{2^{n+2}},\frac{1}{2^n}\right]\\
0 &t\in[0,1]\setminus\left[\frac{1}{2^{n+1}},\frac{1}{2^{n}}\right]
\end{cases}
$$
Show that $d(g_m,g_m)\geq 1$ for $m\neq n$. Conclude that $\{g_n:n\in\mathbb{N}\}$ have no convergent subsequence.
(ii) 
Show that balls $B(g_n, 1/4)$ are contained in $B(0,1)$. Show that there is no finite $1/4$-net to cover this balls, and the consequence there is no finite $1/4$-net for $B(0,1)$.
(iii) Consider the following open cover 
$$\{B(g_n,1/4):n\in\mathbb{N}\}\cup\left\{B(0,1)\setminus\bigcup\limits_{n=1}^{\infty}\overline{B(g_n,1/8)}\right\}
$$
of $\overline{B(0,1)}$. Show there is no finite subcover.
P.S.
There are much more short ways to answer your questions but they are indirect.
