$X$ is a non negative ramdom variable, prove that $EX = \int_{0}^{\infty} P(X>t)dt$.
Well I don't know how to start to prove this, I have a clue that says to me that is: write $P(X>t)$ as the integral of the index function and use the Fubini's theorem.
As you said, you can use Fubini's theorem to write
$$ \int_0^\infty P(X>t)dt = \int_0^\infty E [\mathbb 1_{\{X>t\}}]dt = E \left[\int_0^\infty \mathbb 1_{\{X>t\}}dt \right] = E \left[X \right]. $$
The use of Fubini is justified by the fact that the integrand is non-negative.
Start with the definition and integrate it by parts
$$\mathbb{E}[X]=\int_0^{\infty} [1-F_X(x)]dx= 0 +\int_0^{\infty} xf(x)dx$$
Crucial and simple observation is $X = \int_0^X ds$ if $X$ is non-negative.
Then due to Fubini (non negative):
$$ \mathbb E[X] = \mathbb E[\int_0^X ds] = \mathbb E[\int_0^\infty 1_{\{X>s\}}ds] = \int_0^\infty \mathbb E[1_{\{X>s\}}]ds = \int_0^\infty \mathbb P(X>s)ds$$
