Thank you r9m.
Here is another idea.
Suppose that \(\displaystyle{\mathcal{P}}\) (the set of polynomials) is open in \(\displaystyle{\left(C([1,1]),\cdot_{\infty}\right)}\).
Let \(\displaystyle{f(x)=x\,,x\in\left[1,1\right]}\) and then \(\displaystyle{f\in\mathcal{P}\subseteq C([1,1])}\).
There exists \(\displaystyle{\epsilon>0}\) such that \(\displaystyle{B(f,\epsilon)\subseteq \mathcal{P}}\).
Consider the continuous function \(\displaystyle{g(x)=f(x)+\dfrac{\epsilon\,x}{2\,(1+x)}\,,x\in\left[1,1\right]}\)
Observe that \(\displaystyle{\forall\,x\in\left[1,1\right]\,\,,g(x)f(x)=\dfrac{\epsilon\,x}{2\,(1+x)}<\dfrac{\epsilon}{2}}\)
so, \(\displaystyle{gf_{\infty}\leq \dfrac{\epsilon}{2}<\epsilon\implies g\in B(f,\epsilon)\implies g\in\mathcal{P}}\)
which is a contradiction since \(\displaystyle{g}\) is not twice differentiable at \(\displaystyle{x=0}\).
