about $C([0,1])$ with sup metric Define the space $C([0,1])$ as the space of continuous functions $f : [0,1] \mapsto \Bbb R$ with   $C([0,1])$ $$ d(f,g) = \sup _{x \in [0,1]}{|f(x)-g(x)|} ,  $$
so  let $$A= \left\{f \in C([0,1])\ \middle| \ 0 <\int_0^1 f(x)  \ \mathrm{d}x  < 1\right\}$$  now is $A$ open , close , bounded , connect or compact ? 
I think $A$ is open because for every $f \in A $ we have $B_t (f) \subseteq A $ such that $t:= 1- \int_0^1 f(x)  \ \mathrm{d}x $.(note that $B_t (f) $ is open ball with center $f$ and radios $t)$. $A$ is not close because if we let $f_n (x)= \frac{1}{n}$ then $ 0< \int_0^1 f_n(x)  \ \mathrm{d}x=\frac{1}{n} <1 $ and for every $1< n \in \mathbb{N}$ ,$  f_n(x) \in A$ and $lim_{n \to \infty} f_n(x)=0$  then $  \int_0^1 lim_{n \to \infty} f_n(x)  \ \mathrm{d}x=0  $ then 
 $ lim_{n \to \infty} f_n(x) \notin A$ hence $A$ is not close and yet $A$ is not compact .
 A: The integral function is a linear application and a member of the dual of the space $C([0,1])$ so you have that 
$A=(\int_0^1dx)^{-1}((0,1))$
that it is open because it is the inverse image of an open set with respect to a continuos function.
So $A$ it is not closed because $A\neq C([0,1])$, $A\neq \emptyset$ and  $C([0,1])$ is a topological vector space and so it is connected.
$A$ it is not compact because it is not closed in $C([0,1])$
It is connected because it is a convex subset.
It is not bounded because you can choose for every $\epsilon>0$ a continuos function $f_\epsilon$ such that there exists $x\in [0,1]$ for which $f_\epsilon(x)>\epsilon$ but $0<\int_0^1f_\epsilon(x)dx<1$
A: $A$ is not bounded
The piecewise linear map defined by $f_n(0)=f_n(1/n)=f_n(1)=0$ and $f_n(1/(2n))=n$ is such that $\int_0^1 f_n = 1/2$ but $d(f_n,0) = n$ is unbounded.
$A$ is connected
$A$ is connected because it is convex.
A: The map$$\begin{array}{rccc}I\colon&C\bigl([0,1]\bigr)&\longrightarrow&\mathbb{R}\\&f&\mapsto&\displaystyle\int_0^1f(x)\,\mathrm dx\end{array}$$is continuous. Since $A=I^{-1}\bigl((0,1)\bigr)$, $A$ is open. But $A$ is not closed, by the argument that you used. SInce it is not closed, it is not compact. But it is connected: if $f,g\in A$, just consider$$\begin{array}{rccc}\gamma\colon&[0,1]&\longrightarrow&A\\&t&\mapsto&tg+(1-t)f.\end{array}$$Then $\gamma$ is continuous, $\gamma(0)=f$, and $\gamma(1)=g$. Therefore $A$ is path-connected.
