$l1$ embeds in $X$ Banach space implies $X$ can't be reflexive. Let $X$ and $Y$ be Banach spaces. We say $X$ is isomorphic to $Y$ if there exists a bijective linear bounded operator $T:X\rightarrow Y$ (Note that by the open mapping theorem the inverse is also continuous). 
If $X$ is isomorphic to a subspace of $Y$ we say $X$ embeds into $Y$ or $Y$ contains $X$. 
Every closed linear subspace of a reflexive space is reflexive. $l1$ is not reflexive.
Now, let $X$ be a Banach space such that $l1$ is in $X$, $X$ is not reflexive. 
I can't make sense of the statement, since I don't see reasons to believe that $l1$ inside $X$ is closed. The statement is used to prove that there is no separable reflexive space that contains all other separable reflexive spaces, so you might as well assume that $X$ is separable. Can anyone help me?

It's true that $l_1$ is closed in any $X$ which contains $l_1$. Note that $l_1$ is complete (it is well-known fact). 
Let $\{x_n\}$ be any convergent sequence in $l_1 \subset X$. It's convergent sequence, so it's also Cauchy sequence. $l_1$ is complete, so $\lim_{n \to \infty}x_n=x \in l_1$, so $l_1$ is closed in $X$.
