As the title says the problem is to show that $c_{00}$ is not contained in any maximal ideal ($c_{00}$ is considered lying in $c_0$). I am not used nonunital algebras (rings) so I don't know how to approach to questions like this.
Edit: $c_0 = \{ x_n \in \mathbb{C}, n \in \mathbb{N} : \lim x_n = 0 \}$ and $c_{00}$ is subspace (subalgebra) in $c_0$ with $x_n$ nonzero for only finitely many indexes. I am sorry, I thought the notation was standard.
As remarked in the comments, $c_{00}$ is dense in $c_0$, so it suffices to show that every maximal ideal is closed. This is not true in general commutative Banach algebras.
But by Prop 3.1 and the discussion before that in this paper, we have that every dense maximal ideal $M$ in $c_{0}$ satisfies $c_{0}^2 \subset M$. But $c_{0}^2=c_{0}$, as one can take pointwise square roots of any sequence in $c_{0}$ and the resulting sequence still lies in $c_{0}$., thus having $c_{0}^2 \subset M$ is impossible for a proper ideal.