Let $\mathcal{B}$ be a countable base for the topology.
And let $J \subset I$ be the indexes such that
$O_j \in \mathcal{B}$ for $j \in J$.
First, we will to show that
$$
O = \bigcup_{j \in J} O_j.
$$
It is evident that $\bigcup_{j \in J} O_j \subset O$.
On the other hand, for each $i \in I$, there is a subfamily
$\mathcal{B}_i \subset \mathcal{B}$ such that
$$
O_i = \bigcup_{U \in \mathcal{B}_i} U.
$$
Of course, if $U \in \mathcal{B}_i$, then $u = 0$ a.e. in $U$,
and therefore, $U = O_j$ for some $j \in J$.
That is,
$$
O
=
\bigcup_{i \in I} O_i
=
\bigcup_{i \in I}
\bigcup_{U \in \mathcal{B}_i}
U
\subset
\bigcup_{j \in J} O_j,
$$
as we wanted to show.
Since $J$ is countable (because $\mathcal{B}$ is),
we can conclude that $u = 0$ a.e. in $O$.
It is evident that $O$ is the largest open set such that
$u = 0$ a.e. in $O$.
Therefore, $\mathrm{supp}(u)^c \subset O$.
We have to show that $O \subset \mathrm{supp}(u)^c$.
But this is the same as showing that
$u = 0$ everywhere in $O$.
In order to show this claim, we have to assume that every
non-empty open set has strictly positive measure.
Suppose that $u(x) = \alpha \neq 0$.
Take an open interval $A$ with $x \in A$, and
$0 \not \in A$.
Then, $u^{-1}(A) \cap O$ is an open subset of $O$ where
$u \neq 0$ everywhere.
That is, $u^{-1}(A) \cap O$ has null measure.
From our assumptions, $u^{-1}(A) \cap O$ is empty.
That is, $u = 0$ everywhere in $O$.