Show that this set is open in $E = C([0,1], \mathbb R)$, with the norm $||.||_\infty$ $E = C([0,1], \mathbb R)$, with the norm $||.||_\infty$. Let $O$ be an open of $\mathbb R$ and
$$\Omega(O) = \{ f \in E: f(t) \in O, \forall t \in [0,1] \}$$
Show that $\Omega(O)$ is open in $E$
I proceed as follows: 
$\Omega(O)$ is open in $E$ iff its complement
$$\Omega(O)^c = \{ f \in E: f(t) \in O^c, \forall t \in [0,1] \}$$
is closed in $E$.
Let $(f_n)_{n \geq 1}$ be a convergent sequence in $\Omega(O)^c$ and $f_n \to f$. We need to show that: $f \in \Omega(O)^c$, or equivalently $f(t) \in O^c, \forall t \in [0,1]$
We have: $\forall n \in \mathbb N, \forall t \in [0,1], f_n(t) \in O^c$ closed in $\mathbb R$. $f_n(t) \to f(t), \forall t \in [0,1]$, so we must have: $f(t) \in O^c, \forall t \in [0,1]$. Done.
But we have nothing to do with the norm (?!) Is my proof correct?
 A: Let $f \in \Omega(O)$. So $f(t) \in O, \forall t \in [0,1]$.
We need to show that there exist a ball centered $f$ contained in $\Omega(O)$, 
which means: $\exists \epsilon > 0$: if $||f-g||_\infty < \epsilon$, then $g \in \Omega(O)$.
or, $\exists \epsilon > 0:$ if $|f(t)-g(t)| < \epsilon, \forall t \in [0,1]$, then $g(t) \in O, \forall t \in [0,1]$.
or, $\exists \epsilon > 0: \big (f(t)- \epsilon; f(t) + \epsilon \big ) \subset O, \forall t \in [0,1]$.
Now, as $f([0,1])$ is compact on $\mathbb R$, it attains its maximum and minimum at $a$ and $b$ in $[0,1]$, respectively, so $f(a),f(b) \in f([0,1]) \subset O$. By the openness of $O$, there exist positive $\epsilon_a$ and $\epsilon_b$ such that: $\big (f(a)- \epsilon_a; f(a) + \epsilon_a \big )$ and $\big (f(b)- \epsilon_b; f(b) + \epsilon_b \big )$ are contained in $O$.
Let $\epsilon = \min(\epsilon_a,\epsilon_b)$. It is easy to show that with this $\epsilon, \big (f(t)- \epsilon; f(t) + \epsilon \big ) \subset O, \forall t \in [0,1]$. 
We are done.
A: As mentioned in one of the comment, your proof is not correct. You made an erroneous negation of the statement $f \in \Omega(O)$.
Here is what I propose to do, with a direct proof. Take $f \in \Omega(O)$. We have to find $\epsilon > 0$ such that $B(f,\epsilon)=\{g \in E \, | \, \Vert g-f\Vert < \epsilon \} \subset \Omega(O)$.
As $O$ is open, for all $t \in [0,1]$, you can find $\epsilon_t >0$ such that $(f(t)-2\epsilon_t,f(t)+2\epsilon_t) \subset O$. Now as $f$ is supposed to be continuous, for all $t \in [0,1]$, you can also find $\delta_t >0$ such that for $\vert u-t \vert < \delta_t$, you have $\vert f(u)-f(t) \vert < \epsilon_t$
As $[0,1]$ is compact, you can cover it by a finite number of intervals $(t_i-\delta_i, t_i+\delta_i)$, $1 \le i \le n$. Name $\epsilon = \inf\limits_{1 \le i \le n} \epsilon_i$.
Now suppose that $\Vert g- f \Vert <\epsilon$. For $t \in [0,1]$, there is $j \in \{1, \dots , n\}$ such that $t \in (t_j -\delta_j,t_j+\delta_j)$.
And then $$\vert g(t)-f(t_j) \vert \le \vert g(t)- f(t) \vert +\vert f(t)-f(t_j)\vert \le 2\epsilon_j.$$
Therefore $g(t) \in O$ for all $t \in [0,1]$ which ends the proof.
