Is closed phase trajectory a necessary feature of any one-dimensional periodic motion?

Yes.

• A phase space trajectory of a smooth system$^1$ has to be a continuous curve.

• For it to be called "periodic", the movement has to repeat itself, both velocity and position: i.e., it must come back to the same spot in phase space.

• For a deterministic system, the current condition determines uniquely its future evolution, and there can not be "crossroads".

All of that together mean that, once a phase space trajectory comes back to where it was started, it's forced by determinism to run over the same path again, which forms a closed loop, given continuity.

_{Source: Reartes, W., Actas del XII Congreso Dr. Antonio A. R. Monteiro (2013), 2014, pp. 98$-$103 (PDF file).}

Notice that pendulum rotations can be considered either divergent or periodic. They're divergent when the phase space is the plane, and periodic when it's a cylinder ($\theta = 0$ and $\theta=2\pi$ identified), which means that also rotations are closed curves (over the cylinder) when considered periodic.

As for a formal proof, most texts on ODE's and dynamical systems will have some. For example Hirsch-Smale-Devaney chapters 2 and 3, or the first six pages of Lebovitz textbook chapter on dynamical systems.

_{$^1$ Dr Xorile's answer shows this condition can be relaxed.}

No. At least if I understand your question correctly.

Say, a simple harmonic oscillator has a $\cos$ position curve. Then the velocity curve is going to be $dp/dt$ which will be $\sin$, and so you get your ellipse when you plot them on a phase diagram.

But now consider saw-tooth motion. Then the velocity curve switches from a constant positive, to a constant negative ($dp/dt$ is constant, because the saw-tooth is simply a straight line).

If you plot position against velocity, you end up with a box (or two parallel lines, with an instantaneous jump between them). So closed figure, but not an ellipse.

Furthermore, while a closed loop is necessary, there's no reason that the loop needs to be simple. It can overlap with itself. @stafusa's answer adds the condition that the path be deterministic, but this condition, too, can be relaxed.

Considered constant speed motion moving from $-2\rightarrow1\rightarrow-1\rightarrow2\rightarrow2$. Here the phase diagram is a box that does a kind of loop on its way over.

Specifically, it's a path of straight lines going from (in $(p,v)$ space):

$(-2,1)\rightarrow(1,1)\rightarrow(1,-1)\rightarrow(-1,-1)\rightarrow(-1,1)\rightarrow(2,1)\rightarrow(2,-1)\rightarrow(-2,-1)\rightarrow(-2,1)$

Note that the line segment from $(-1,1)\rightarrow(1,1)$ is covered twice in this loop.