The well-known art gallery problem starts with an "art gallery" (a simple polygon in the plane, not necessarily convex) and asks for the minimum number of "guards" (points on the polygon) required to "observe the whole gallery" (to have the property that for any point in the interior of the polygon, there is a line segment from that point to a "guard" that lies entirely within the polygon). Chvatal showed that if the polygon has $n$ vertices, then $\lfloor n/3\rfloor$ guards are sufficient, and sometimes necessary, to observe the whole gallery.
If you forget about trying to minimize the number of guards, and simply want to place guards so that they see the whole gallery, it is reasonably clear that if you place a guard at each vertex of a simple polygon, they will be able to observe the whole gallery.
One way to see this is to note that the assertion is clear for triangles, and then to recall (or to convince oneself) that any simple polygon can be triangulated without adding vertices.
If visualizing an entire triangulation of the polygon is too "global", one can think "locally" as follows. Fix any point $p$ in the gallery interior. Choose a point $q$ on the polygon such that the distance from $p$ to $q$ is minimized. The line segment from $p$ to $q$ lies within the gallery. If $q$ is a vertex, we are done. Otherwise, $q$ is on the interior of an edge. Pick a direction on that edge and move $q$ along the edge in that direction. Eventually, one of two things will happen: (1) the point $q$ becomes a vertex, or (2) there is a first time at which the line segment from $p$ to $q$ intersects the polygon somewhere besides $q$. In case (1) we are done, and in case (2), we can convince ourselves that at the time that this happens, the closest point to $p$ on the intersection of the polygon and the line segment must be a vertex, and we are again done.
Now switch from two dimensions to three so an "art gallery" is now a polyhedron. If you place a guard at each vertex, can they observe the whole gallery?
The answer, in general, is no.
It may not be clear why it is no, but it is relatively clear that the arguments just given do not generalize in any simple way.
There are polyhedra that cannot be "triangulated" into tetrahedra without adding vertices. A famous example of this is the Schoenhardt polyhedron. (Yet: experimenting with this applet convinced me that the vertices of this polyhedron do see all of its interior.)
The "given $p$, pick a closest point $q$ on the polyhedron, and then move $q$ in some direction" idea clearly cannot work (at least without judicious choice of direction), because in the case (2) there is no reason for the closest point on the intersection of the line segment from $p$ to $q$ with the polyhedron to be a vertex in the three-dimensional case. It can pretty obviously be on the interior of some edge.
So it's not counterintuitive, to me, that there are polyhedra whose vertices cannot observe their interiors. But I'd like a better mental image of what such a polyhedron can actually "look like." (A better image, for example, than what I get from the picture on the Wikipedia entry for the art gallery problem.)
Can somebody describe a polyhedron, in such a way that it is in some sense "obvious" that its vertices cannot see all of its interior? So that it is possible to form a clear mental picture of what it would look like to be inside such a polyhedron, at a point where you can't see the vertices? (What do you see?)