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The following gives a partial answer: no such unbounded connected set may exist with the further assumption that it is closed. Actually, the argument generalizes for any locally compact metric space. I'm not completely sure that a much simpler or even trivial proof may exist, though.

Let $\Gamma$ be a closed unbounded connected subset of the plane. Let $x\in\Gamma$ and let $B:=B(x,r)$ be an open ball around $x$. I claim that the connected component of $x$ in $\Gamma\cap \bar{B}$ meets $\partial B$, which shows that $\Gamma$ does contain non-trivial bounded connected subsets.

For any $\epsilon>0$, consider the $\epsilon$-neighborhood of $\Gamma,$ that is $\Gamma_\epsilon:=\cup_{y\in\Gamma}B(y,\epsilon).$ It is an open unbounded connected subset of the plane. Let $U_\epsilon$ be the connected component of $x$ in $\Gamma_\epsilon\cap B$. Since the latter is locally connected, $U_\epsilon$ is both an open and closed subset of it in the relative topology. It is therefore an open subset of $\Gamma_\epsilon$; however it is not closed in it, because $\Gamma_\epsilon$ is connected. Therefore $\bar U_\epsilon$ is a closed connected set that meets $\partial B,$ and of course contains $x$. Since the set of all connected closed subsets of a compact metric space is compact in the Hausdorff distance, taking a limit as $\epsilon\to0$ we get a bounded connected subset of $\Gamma$ connecting $x$ with $\partial B$ (this also passes to the limit).

Rmk One could state the above in terms of the one-point compactification of $\Gamma$, and more generally for compact connected metric spaces. The trick of approximating a metric space with a locally connected metric space is made possible via the Kuratowski embedding (one defines $X_\epsilon$ as an $\epsilon$ nbd of $X$ in the embedding).

PS: Of course the same affirmative conclusion holds, even more directely, if $\Gamma$ is assumed to be open, which is another case included in the original assumption of completely metrizable.

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The following gives a partial answer: no such unbounded connected set may exist with the further assumption that it is closed. Actually, the argument generalizes for any locally compact metric space. I'm not completely sure that a much simpler or even trivial proof may exist, though.

Let $\Gamma$ be a closed unbounded connected subset of the plane. Let $x\in\Gamma$ and let $B:=B(x,r)$ be an open ball around $x$. I claim that the connected component of $x$ in $\Gamma\cap \bar{B}$ meets $\partial B$, which shows that $\Gamma$ does contain non-trivial bounded connected subsets.

For any $\epsilon>0$, consider the $\epsilon$-neighborhood of $\Gamma,$ that is $\Gamma_\epsilon:=\cup_{y\in\Gamma}B(y,\epsilon).$ It is an open unbounded connected subset of the plane. Let $U_\epsilon$ be the connected component of $x$ in $\Gamma_\epsilon\cap B$. Since the latter is locally connected, $U_\epsilon$ is both an open and closed subset of it in the relative topology. It is therefore an open subset of $\Gamma_\epsilon$; however it is not closed in it, because $\Gamma_\epsilon$ is connected. Therefore $\bar U_\epsilon$ is a closed connected set that meets $\partial B,$ and of course contains $x$. Since the set of all connected closed subsets of a compact metric space is compact in the Hausdorff distance, taking a limit as $\epsilon\to0$ we get a bounded connected subset of $\Gamma$ connecting $x$ with $\partial B$ (this also passes to the limit).

Rmk One could state the above in terms of the one-point compactification of $\Gamma$, and more generally for compact connected metric spaces. The trick of approximating a metric space with a locally connected metric space is made possible via the Kuratowski embedding (one defines $X_\epsilon$ as an $\epsilon$ nbd of $X$ in the embedding).