Assume $\Omega$ is an open set in $\mathbb R^3$
such that the intersection of $\Omega$ with any horizontal plane is simply connected. 



> 
> Can you prove that $\Omega$ is simply connected?
> 
> 
> 


(Note that by the definition, simply connected set can not be empty.)


**Comments.**


* The proof given by Tom Goodwillie below is done with bare hands. I would prefer to find ready to use tool for answering this and similar questions.



Simply connected slices
Yes, I think so. Let's show that every compact set $K\subset \Omega$ is contained in some compact contractible subset of $\Omega$. We use the fact that in a simply connected open subset of the plane every compact set is contained in some compact contractible set.


Denote by $P\_t$ the plane $\mathbb R^2\times t$, and define the set $\Omega\_t\subset\mathbb R^2$ by $\Omega\_t\times t=\Omega\cap P\_t$. Define $K\_t$ likewise.


For each $t$ choose a compact contractible set $C\_t\subset \Omega\_t$ such that $K\_t\subset C\_t$. There must be an interval $J\_t$ containing $t$ such that for every $t'\in J\_t$ we have $K\_{t'}\subset C\_t\subset \Omega\_{t'}$.


The set of all $t$ such that $K\_t$ is nonempty can be covered by finitely such intervals. Thus for some $a$ there are real numbers $s\_0<\dots <s\_a$ and numbers $t\_i\in [s\_{i-1},s\_i]$ such that 
$$
K\subset \cup\_{i=1}^a ([s\_{i-1},s\_i]\times C\_{t\_i})\subset \Omega.
$$
Enlarge this union to make it contractible by choosing, for each $i=1,\dots a-1$, a compact contractible set $D\_i$ such that $C\_{t\_i}\cup C\_{t\_{i+1}}\subset D\_i\subset \Omega\_{t\_{i-1}}$ and then adding the sets $s\_i\times D\_i$.


