Does $L^1$ contain a subspace isomorphic to $c_0$?

This is the proof that I know. It's not easy, but it is well known in the literature. $L^1$ has a property called cotype 2, that is, there is a constant $C>0$ such that for any $x_1,\dots,x_n \in L^1$ $$ E \left\|\sum_{k=1}^n \epsilon_k x_k \right\|_1 \ge C \left(\sum_{k=1}^n \|x_k\|_1^2\right)^{1/2} ,$$ where $\epsilon_k$ is a sequence of independent identically distributed random variables satisfying $\Pr(\epsilon_k = 1) = \Pr(\epsilon_k = -1) = 1/2$. This can be proved using Khintchine's inequality.

This property will be inherited by any subspace, and carries over to isomorphisms.

The space $c_0$ does not have this property. The easiest way to see this is to consider the case that $x_k$ are the unit vectors.

Here's another proof:

A non-reflexive subspace of $L_1$ contains an isomorphic copy of $\ell_1$. This follows from the Dunford-Pettis characterization of weakly compact subsets of $L_1$. c.f., Kalton and Albiac, Topics in Banach Space Theory, Theorem 5.2.9. and Proposition 5.6.2; or Theorem 8 from these notes by Joe Diestel.

On the other hand, $c_0$ does not contain $\ell_1$ isomorphically (otherwise, $c_0$ would have non-separable dual).