= Solution
The <Baire category theorem> states that a complete metric space is not a countable union of closed sets with empty interior. To prove it, let $G_1,G_2,\ldots$ be open dense sets and start with any nonempty <open set> $V$. Inductively choose closed balls
$$
\overline B(x_n,r_n)\subset
B(x_{n-1},r_{n-1})\cap G_n,
\qquad 0<r_n<2^{-n},
$$
with the first ball inside $V\cap G_1$. The centres are Cauchy. Completeness supplies a <limit> lying in every ball, hence in $V\cap\bigcap_nG_n$. Thus the intersection is dense, which is the equivalent form of the theorem.
Now each $nS$ is closed and $\bigcup_{n\geq1}nS=X$. Baire gives
$$
B(x,r)\subset nS
$$
for some $x,r,n$. Symmetry gives $B(-x,r)\subset nS$. If $\|z\|<r$, then $x+z$ and $-x+z$ lie in these two balls; convexity makes their midpoint $z$ belong to $nS$. Hence
$$
B(0,r/n)\subset S,
$$
which proves the <closed convex absorbing set has an origin neighbourhood> result.
Convexity cannot be dropped. In $X=\mathbb R$, set
$$
S=\{0\}\cup\bigcup_{k\geq0}
\{x:4^{-k}\leq|x|\leq2\cdot4^{-k}\}.
$$
This is closed and symmetric but has gaps arbitrarily close to zero. For any $x\ne0$, choose $k$ so that the interval
$$
\left[\frac{|x|}{2\cdot4^{-k}},\frac{|x|}{4^{-k}}\right]
$$
has length at least one, and choose an integer $n$ in it. Then $x/n\in S$, so $\bigcup_n nS=\mathbb R$, while $S$ is not a neighbourhood of zero. This is a <closed symmetric absorbing set without an origin neighbourhood>.
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