If a set is open, its "slices" are also open. Converse?

This is almost always false. To put this in a perhaps more familiar context, note that the collection of all sets $U\subseteq X\times Y$ whose slices are open is a topology; call it $T$. Now note that a map $f:X\times Y\to Z$ is continuous with respect to $T$ iff it is continuous on each coordinate separately: that is, iff for each $x\in X$, $y\mapsto f(x,y)$ is continuous, and for each $y\in Y$, $x\mapsto f(x,y)$ is continuous. So your question is equivalent to asking whether every separately continuous function on a product is actually jointly continuous (i.e., continuous with respect to the product topology).

As multivariable calculus students learn, this is false even in the most familiar contexts. For instance, taking $X=Y=Z=\mathbb{R}$, the function $f(x,y)=\frac{xy}{x^2+y^2}$ for $(x,y)\neq (0,0)$ and $f(0,0)=0$ is continuous in each variable separately but is not continuous jointly at $(0,0)$. To get a direct counterexample to your question, you can take $U=f^{-1}(\mathbb{R}\setminus\{1/2\})\subset\mathbb{R}^2$. This set has open slices, but it is not open in the product topology (it contains $(0,0)$ but does not contain any ball around $(0,0)$ since $f(a,a)=1/2$ for any $a\neq 0$). Even more explicitly, this set $U$ is the complement of the set $\{(a,a):a\neq0\}$, which you can easily verify has closed slices but is not closed