Let $(X,\tau)$ be connected such that $\emptyset$ and $X$ are the only open connected subsets. Does this imply that $\tau = \{\emptyset, X\}$?

1$\begingroup$ I am thinking this is simple, but I could not give an Yes/No answer.. D: I am over simplifying the question I guess... :D Can you please tell me why this question is not simple.. $\endgroup$ – Praphulla Koushik Dec 22 '18 at 15:57

$\begingroup$ I can't say for sure that this is nontrivial... It has happened to me that I have asked utterly trivial questions on MO and had to remove them. Hopefully, this question here is somewhat interesting. $\endgroup$ – Dominic van der Zypen Dec 22 '18 at 16:35

$\begingroup$ If U is a proper and nonempty subset of X which is open, then the complement of U in X is not open, which suggests the topology may be an ultrafilter or a subset of one. Have you considered a non principal ultrafilter as a candidate for a topology? Gerhard "It's One Or The Other" Paseman, 2018.12.22. $\endgroup$ – Gerhard Paseman Dec 22 '18 at 17:07

$\begingroup$ Maybe edit the title to prepend "Nontrivial" $\endgroup$ – Neal Dec 22 '18 at 17:26

1$\begingroup$ At this point, the question has been completely answered. I just want to highlight two simple facts. (1) If $X$ is a finite topological set, then any nonempty open subset of minimal cardinality is connected as it cannot split into smaller nonempty open subsets. Thus the implication holds. (2) We can build for any infinite set $X$ a nontrivial countable topology $\tau$ with the desired property. Indeed, take $Y$ an infinite proper subset of $X$ and write $Y = Y_0 \cup Y_1$ where the $Y_i$ are disjoint and infinite. Then iterate by splitting each set $Y_i$ to get $\tau = \{0, X, Y_w \}$. $\endgroup$ – Luc Guyot Dec 22 '18 at 20:12
Let $X_0$ be any topological space in which no nonempty open set is connected, e.g., $X_0 = \mathbb{Q}$ with the usual topology. Then let $X = X_0 \cup \{\infty\}$ with open sets $\{$all open subsets of $X_0\} \cup \{X\}$. The whole space is connected because the only open set containing $\infty$ is the whole space, and any proper nonempty open subset cannot contain $\infty$ and hence must be disconnected by the assumption on $X_0$.
This is false.
Example. Let $Y$ be a totally disconnected topological space in which no point is open; for example $Y = \mathbb Z_p$. Construct $X$ from $Y$ by adding a closed point in the closure of every nonempty open: $X = Y \cup \{x\}$, where $U \subseteq X$ is open if and only if $U = X$ or $U \subseteq Y$ open.
Because the only open set containing $x$ is $X$, we see that every open cover of $X$ has to contain $X$ as one of its opens, so in particular $X$ is connected.
On the other hand, every strict open $U \subsetneq X$ is contained in $Y$, hence is totally disconnected. Since points are not open by assumption, $U$ is disconnected. $\square$

3$\begingroup$ I would like to accept both your and Nik Weaver's answer, as they provide similar and interesting arguments, but Nik's answer was first, so I accept his answer, and +1 both of your answers. I hope you understand $\endgroup$ – Dominic van der Zypen Dec 22 '18 at 17:52