Show quotient topology on [0,1] = usual topology on [0,1]

Let X = [-1,1], and give X the usual topology.

Definte a function f:X -> [0,1] by f(x) = x^2. Show that the quotient topology on [0,1] induced by f is the same as the usual topology on [0,1].

This is so tricky. I figured out that I kinda have to prove that both are subsets of each other. So I need to say that the open subsets of f^-1(u) in [-1,1] lead to an open set in the usual sense of [0,1]. Then I need to say that the open subsets in [0,1] have open pre-images of f in X.

Trouble is, it's impossible to generalize an open subset. I think it's something like two half-open intervals at the end of the subset, with a union of open intervals in the middle?

Any help will be GREATLY appreciated.