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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 $X$?

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There is no such space. For if $x$ is any point and $X\setminus \{x\}=U|V$ 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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    $\begingroup$ What is $X\setminus \{x\}=U|V$? $\endgroup$
    – Rafał Gruszczyński
    Commented Nov 14, 2017 at 7:33
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    $\begingroup$ I've never seen this notation -- but I like it. $\endgroup$
    – Jochen Wengenroth
    Commented Nov 14, 2017 at 9:39
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    $\begingroup$ A standard notation for disjoint unions is $U\sqcup V$ ($\backslash$sqcup) $\endgroup$
    – YCor
    Commented Nov 14, 2017 at 10:59
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    $\begingroup$ @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
    Commented Nov 14, 2017 at 11:01
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    $\begingroup$ 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
    Commented Nov 14, 2017 at 13:28

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