I am trying to find a graph (ideally the smallest) that demonstrates why the following 3 necessary conditions are not sufficient for a graph to be traceable. In other words, what is the smallest untraceable graph $G$ satisfying the following 3 conditions?
- $G$ is connected.
- $G$ has at most two vertices of degree 1.
- For all $S \subseteq V(G)$, $G - S$ has at most $\vert S \vert + 1$ components.
Wolfram gives drawings of all connected untraceable graphs of order 6 and below but none of these satisfy condition 3.
This has 7 vertices. You've found 6 is not enough. In general, having lots of small cuts (in this case, 3 cut vertices) can help with generating these sorts of counterexamples.