A linear problem is always convex, because anything linear is convex.
As pointed out by @Marco Lübbecke, any linear function is also concave. But polygons (feasible sets of linear programs) are only convex (and not concave).
Check out this link, it is well explained, or this one for an algeabraic proof.
Your example has only one feasible point (assuming $x$ and $y$ are positive) : $(0,3)$. I suspect you were maybe thinking of an example such as $y\le 1$ OR $y\ge 2$. This indeed is not convex. Both constraints are linear, but the OR operations kills the convexity.