Open sets in a topology produced by metrics. For each $i\in I$, $$d_i:X^2\to \Bbb R$$ is a metric and $\mathcal T$ is the coarsest topology on $X$ containing all topologies produced by the $d_i$ .
For $U\subseteq X$ we have:
$$\forall x \in U,~\exists i\in I,~\exists r>0,~~B_x^{d_i}(r)\subseteq U$$
where $B_x^{d_i}(r)$ is the open ball with center $x$ and radius $r$.
Is $U$ open in $(X,\mathcal T)$?

Edit:
Is $U$ open in some $(X,d_i)$?

Yes. For any $x\in X,i\in I$ and $r>0$, $B_x^{d_i}(r)\in\mathcal{T}_i$, the topology generated by $d_i$, and $\mathcal{T}_i\subseteq\mathcal{T}$, so $B_x^{d_i}(r)\in\mathcal{T}$ for each $x\in X,i\in I$ and $r>0$. In particular, each point of $U$ has a $\mathcal{T}$-open nbhd contained in $U$, so $U\in\mathcal{T}$.
