Let $E$ be a Topological Vector Space and $U$ a bounded set of $E$ with $0\in U$, i.e. given any neighborhood $W$ of the origin, there exist $\alpha>0$ such that $\alpha U\subset W$. Is it true that $$\bigcap_{n=1}^\infty2^{-n}U=\{0\}$$
Note: I have added the hypothesis $0\in U$.
Nothing wrong with Hagen's solution, I just wanted to complete my suggestion to a full solution starting from first principles :-).
I assume that $E$ is Hausdorff. I argue that this implies that the intersection of those scaled versions of $U$ cannot contain a non-zero vector. Let $v\in E, v\neq0$. Then there exists an open neighborhood $V_1$ of the origin with the property that $v\notin V_1$. By continuity of the scalar multiplication there exists another neighborhood $V_2$ of the origin and an open interval $I_a=(-a,a),a>0,$ such that for all $x\in V_2, \alpha\in I_a$ we have $\alpha x\in V_1$. Let $V_3=(a/2)V_2$. Multiplication by $a/2$ is a homeomorphism, so $V_3$ is an open neighborhood of the origin. I claim that for all $t\in(0,1]$ we have $tV_3\subseteq V_1$. So let $t\in(0,1]$ and $x\in V_3$ be arbitrary. We have $x=(a/2)y$ for some $y\in V_2$. Then $tx=(ta/2) y\in V_1,$ because $(y,ta/2)\in V_2\times I_a$.
Now let $$ W=\bigcup_{t\in(0,1]} tV_3.$$ This is an open neighborhood of the origin as a union of open sets. We just checked that $W\subseteq V_1$, so $v\notin W$. Furthermore $W$ has the property that $tW\subseteq W$ for all $t\in(0,1]$. Finally we are in a position to attack the main claim. By boundedness of $U$, we have $\alpha U\subseteq W$ for some $\alpha>0$. If $2^{-n}<\alpha$, then $t=2^{-n}/\alpha\in(0,1]$, and therefore $$ 2^{-n}U=t\alpha U\subseteq tW\subseteq W. $$ Hence $v\notin 2^{-n}U$, and the claim follows.