I add this answer to provide some variety, I believe this construction is collapse-resistant but the proof is not as easy and elegant as the other answers. Please comment if you find an error or gap in the proof !

Let $\mathbb{G}_f$ be the set of all finite graph. For $g \in \mathbb{G}_f$ a finite graph, let $\hat{g}$ be the union of a countable number of copy of $g$. Now let $G$ be the union of all $\hat{g}$ for $g \in \mathbb{G}_f$.

$G$ is not connected but it is collapse-resistant. When we collapse a finite subset of $G$, the resulting vertex will be connected to a finite number of vertices and be part a finite graph. It can be mapped to the corresponding finite graph from $G$. There will be a countable number of every finite graph remaining (not taking part in the collapse), so the result is still a countable number of every finite graph.

To get a connected graph, let us construct $\bar{G}$ by adding a vertex $o$ to $G$, connected to every other vertex.

$\bar{G}$ is connected and also collapse-resistant. If we collapse a subset $S$ containing $o$, the resulting vertex will be connected to all other vertices and can be mapped to $o$. For the remaining vertices ($\bar{G} \setminus \{o\}$), it is just as if we deleted $S \setminus \{o\}$, it will still result in a countable number of every finite graph. If we collapse a subset not containing $o$, it will work just the same as for $G$ (but with an added vertex $o$ connected to everything)

Note that this graph $\bar{G}$ is not the Rado graph, for example it has a vertex connected to every other vertex.

noedges be collapse-resistant, for the same reason that an infinite complete graph is? $\endgroup$6more comments