$L$ need not be closed. Consider a vector field $X$ with the following properties:
- $X(x,0)=(0,x)$ for all $x>0$.
- For $x\in (0,1)$, the trajectory starting at $(x,0)$ follows a vertical line until it passes through the point $(0,\frac 1{1-x})$, and then travels counter-clockwise around the origin until it returns to $(x,0)$.
- The trajectory starting at $(1,0)$ never leaves the vertical line $x=1$.
- The trajectory starting at $(x,0)$ for any $x>0$ is non-periodic.
Then $L=[0,1)$ is not closed.
In the other direction, given any closed $L$, it is possible to construct a vector field realising that $L$. Indeed, begin by constructing a map $f\colon [0,\infty)\to [0,\infty)$ such that $L = \{ x \mid f(x)=x \}$, and then construct a flow whose first return map to the positive $x$-axis is the map $f$.
To construct the map, just observe that the complement of $L$ is a countable union of open intervals, and so you can define $f(x) = x$ for all $x\in L$, and then define $f$ on each open interval $(a,b)$ as a "north-south" map, that is, a map of the interval $[a,b]$ that fixes $a$ and $b$ and moves every other point towards $b$.
To construct the flow from the map, just draw circles of radius $x$ for every $x\in L$, and then spirals connecting $(x,0)$ to $(f(x),0)$ for every $x\notin L$.
This still leaves open the question of which non-closed sets can be realised. Clearly there's a topological obstruction to using the procedure at the beginning of my answer to remove lots of boundary points from $L$; indeed, once you remove a single boundary point that way, then the trajectory starting at any larger value of $x$ cannot wind around the origin, and so life becomes a little more complicated...