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 .

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.
