Some remarks, not an answer.

1) As pointed out by BS, the problem is: does there exist a connected Polish space containing no nondegenerate proper closed connected subspace? In the language of continuum theory this sounds suspiciously simple: does there exist a connected Polish space whose all composants are singletons? Or (still equivalently): does there exist a connected Polish space that is irreducible between every pair of distinct points? Surely continuum theorists have thought about this question, haven't they?

2) As hinted by Garabed, such a space $X$ cannot be locally compact. Indeed, in locally compact spaces, components coincide with quasi-components. Let $F$ be a closed ball of some radius about some $x\in X$, so that $F\ne X$, and let $C$ be the component of $x$ in $F$.
If $C\ne\{x\}$, then $C$ is a closed connected nondegenerate subset of $X$. If $C=\{x\}$, then $\{x\}$ is also a quasi-component, so $x$ is contained in arbitrarily small clopen sets in $F$. Then they are also clopen in $X$, so $X$ cannot be connected.

3) There exists a connected Polish space $X$ containing no nondegenerate *compact* connected subspace. For instance, the graph of $f(x)=\sum_{n=1}^\infty 2^{-n}\sin(\frac1x-r_n)$, where $r_n$ is the $n$th rational number (in some order) and $\sin(\infty)=0$. See Kuratowski's "Topology" (volume II, section 47.IX in the 1968 edition).
In fact, such an $X$ can even be locally connected (of course, it cannot be locally path-connected at any point).

4) The graph of a discontinuous function $f:\Bbb R\to\Bbb R$ satisfying $f(x+y)=f(x)+f(y)$
can be connected, and in that case it contains no nondegenerate *bounded* connected subset.

Classical Descriptive Set Theory. – François G. Dorais♦ May 19 '10 at 1:36