Suppose $(0,0)$ (or any other point on the x-axis) has a compact neighbourhood $\mathcal{N}$. Then $\mathcal{N}$ contains a basic open neighbourhood $\mathcal{U}$, say $\mathcal{U}$ is $(0,0)$ along with the open disc of radius $r$ centred at $(0,r)$. Since the Moore plane is Hausdorff and $\mathcal{N}$ is compact, $\mathcal{N}$ must be closed. Thus $\mathcal{N}$ contains the boundary $\operatorname{bd}(\mathcal{U})$ of $\mathcal{U}$. Thus $\operatorname{bd}(\mathcal{U})$ is also compact. But $\operatorname{bd}(\mathcal{U})$ is the circle radius $r$ centred at $(0,r)$ with the point $(0,0)$ deleted. In particular, it's contained in the complement of the x-axis, which is homeomorphic to the upper half-plane with the usual topology. But of course a circle with a point deleted is not compact in the usual topology.