Is there a connected $T_2$ space $(X,\tau)$ with more than one point, such that the singletons and $X$ are the only connected subspaces of $X$?
There is no such space. For if $x$ is any point and $X\setminus \{x\}=UV$ then $\{x\}\cup U$ and $\{x\}\cup V$ are connected sets, each with more than one point and different from $X$.
The closest thing you can get is a connected set whose connected subsets are cofinite. The axiom CH implies there is a countable connected Hausdorff space with this property here that is not too complicated (relative to the completely regular example). I don't think a Hausdorff example has ever been constructed in ZFC.

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$\begingroup$ Assuming $X\setminus \{x\}$ is not connected (otherwise the space fails to have the desired property), it is equal to the union of two disjoint nonempty open sets $U$ and $V$. $\endgroup$ – D.S. Lipham Nov 14 '17 at 7:35

2$\begingroup$ A standard notation for disjoint unions is $U\sqcup V$ ($\backslash$sqcup) $\endgroup$ – YCor Nov 14 '17 at 10:59

7$\begingroup$ @erz If $U\cup\{x\}$ is not connected, write it as $W\sqcup Y$ with $x\in Y$, $W,Y$ both nonempty and closed in $U\cup\{x\}$. Then $X=W\sqcup (Y\cup V)$ and both are closed and nonempty. $\endgroup$ – YCor Nov 14 '17 at 11:01

1$\begingroup$ Incidentally, this shows that every infinite connected Hausdorff space admits an infinite connected proper subset which is either open (as the complement of a singleton) or closed. Iterating, we obtain a strictly decreasing sequence of connected subsets, each of which has finite complement in its closure. $\endgroup$ – YCor Nov 14 '17 at 13:28