Assume $(X_i,\mathcal{T}_i)$, $i \in I$ is a family of topological vector spaces and $X:=\prod\limits_{i\in I} X_i$ with the product topology $\mathcal{T}$. $I$ is infinite and for all $i\in I: X_i\neq \{0\}$.
I want to prove that there is no bounded neighbourhood of $0$ in $(X, \mathcal{T})$.
My idea so far was to prove this statement indirectly and assume there is a neighbourhood $U$ which is bounded, i.e. for all 0 neighbourhoods $V$ there is a scalar $\lambda$ such that $U\subset \lambda V$.
Further I know that $V = \prod\limits_{i\in I}U_i$ with $U_i=X_i$ for almost all $i \in I$. Hence there are $i_1,...,i_m \in I$ such that $V = \bigcap\limits_{j=1}^m \pi_{i_j}^{-1}(U_{i_j})$.
Can someone give me a hin how to proceed or lead me to the right direction?
You cannot assume that $V$ is of this form, because those open sets are only a basis for the product topology. $V$ is an union of open sets of this form.
Anyway I think this is no need to make a proof by contradiction : take $U$ a neighborhood of $0$ in the product, then it contains an open set of the basis $\prod_{i\in I}U_i$, which itself contains $0$, and with $U_i=X_i$ for all but a finite number of $i$. Then, there is some $i_0\in I$ such that $U_{i_0}=X_{i_0}$ because the collection is not finite. Now, take $B_0$ an open neighborhood of $0$ in $X_{i_0}$ which is different from $X_{i_0}$, this is possible because $X_{i_0}\neq 0$. Then, form $V=\prod_{i\in I}V_i$ where $V_i=X_i$ except for $i=i_0$, where we take $V_{i_0}=B_0$. Then, $V$ is an open neighborhood of $0$ in the product, and for $\lambda>0$ we have that $p_{i_0}(\lambda V)=\lambda B_0$, while $p_{i_0}(U)=X_{i_0}$, where $p_{i_0}$ is the projection on the $i_0$-th coordinate. So, this is not possible that $U\subset \lambda V$.