Let $A$ be a convex, absorbing subset of a real Banach space $X$ with the additional property that the closure $\rm{cl}(A)$ contains an open ball around $0\in X$.
Does this imply that already $A$ itself contains an open ball around 0?
(I hope that this additional property is already a consequence of Baire's lemma. But please correct me if this is wrong...)
$X$ is a complete metric space, so it is of second category. In this thread is proved that $\overline{A}$ contains a neighborhood of $0$.
No. Let $X=\ell^2$ and $A_0$ be a set consisting of standard unit vectors $e_n$, $n\in\Bbb N$, $A_1$ be a maximal linear independent subset of unit length vectors containing $A_0$, and $A$ be a convex hull of the set $A_1\cup (-A_1)$. Then each vector $x\in X$ has a unique representation as a linear combination of vectors of $A_1$ and $A$ is an absorbing set of $X$. Now for each natural $n$ put $x_n=\tfrac 2n\sum_{i=1}^n e_i$. Then $x_n\not\in A$, but $\|x_n\|=\tfrac 2{\sqrt{n}}$, so a sequence $\{x_n\}$ converges to zero.