Let $X\times Y$ be a topological space (if it helps anything, it can be assumed to be a smooth manifold and $U\subseteq X\times Y$ an open subset.
Let $x\in X$ and $O\subseteq Y$ be open, such that $\{x\}\times O\subseteq U$. Does there exist an open $V\subseteq X$ with $x\in V$ and $V\times O\subseteq U$?
This above question got a negative answer. Let me ask instead the following:
Let in addition $C\subseteq Y$ be a closed set, such that $C\subseteq O$. Does there exist an open $V\subseteq X$ and an open $C\subseteq O'\subseteq Y$ with $x\in V$ and $V\times O'\subseteq U$?
Consider $X=Y=[0,1]$ let
$$U=\{(x,y)\in X\times Y : x+y< 1/2\}.$$
Then $\{0\}\times (0,1/2)\subseteq U$, however any open set around $\{0\}$ in $X$ contains an interval, $[0,b)$ and if $[0,b)\times (0,1/2)\subseteq U$ we can note that this implies $(b,{1-b\over 2})\in U$ which is impossible as $(1-b)/2+b=(1+b)/2>{1\over 2}$.
The modified question is pretty much trivial. Since you don't mind if we assume a smooth manifold, I will. Then the basic separation axioms say that every point has a basis of compact neighborhoods (i.e. compact--hence closed--sets containing an open set). So if you pick any point in $\{x\}\times O$, hence in $U$, we can find a closed set containing an open set which is totally contained in any open set that the point is contained in. Then by the definition of the product topology, we have that all open sets contain a set of the form $A\times B$ with $A,B$ open in $X,Y$ respectively.