Real Analysis, continuity on [0,1]

Q: let f be continuous from [0,1] to [0,1]. Show their is an x on [0,1] for which f(x) = x

Our hint was to use g(x) = f(x) - x and use the IVT

We figure that since our Range is from [0,1] with our max = 1 and min = 0 that if we could show that the function attains a value above y=x and one below y=x, that continuity implies that it mus cross y=x attaining our value.

Problem is how do we do this, my intial thinking was using the defintition of continuity using our

|f(x) - f(a)| < epsilon where we take 2 cases where f(a) is our min and f(a) is our max, where working it down would take us to

|f(x) - f(max)| < epsilon

|f(x) - f(min)| < epsilon

|f(x) - 1| < epsilon

|f(x) - 0| < epsilon

putting our x

|f(x)| < x < |f(x) - 1|

hence where I stopped and got stuck. Thanks for any help