Does $L^1$ contain a subspace isomorphic to $c_0$? Can any $L^1$ space, say $L^1(\mathbb{R})$, have some subspace isomorphic to $c_0$? I guess not but I don't see an argument right now. 
 A: 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.
A: 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).  
