In your system, the dynamics of $x$ and $y$ are not coupled. For a solution $(x(t),y(t))$ to be periodic in $t$, obviously both $x$ and $y$ need to be periodic. Therefore, it suffices to answer the question for the one-dimensional ODE $\dot{x} = f(x)$.
If $x(t)$ solves the ODE $\dot{x} = f(x)$ with $f$ continuous, $x(t)$ is at least continuously differentiable in $t$. Suppose $x_p(t)$ is periodic with period $T$, i.e. $x_p(0) = x_p(T)$. By the mean value theorem, we know that there is a time in between where $\dot{x}_p$ vanishes, i.e. there exists $t_* \in (0, T)$ such that $\dot{x}_p(t_*) = 0$.
Since $x_p$ is a solution of $\dot{x} = f(x)$, we know that the $x_p$-value at $t = t_*$ is a fixed point of $f$, because $\dot{x}_p(t_*) = 0 = f(x_p(t_*))$. However, we also know that the constant solution $x_c = x_p(t_*)$ solves the ODE $\dot{x} = f(x)$, because $\dot{x}_c = 0 = f(x_c)$.
To summarize, we have two solutions of the ODE $\dot{x} = f(x)$ which pass through the point $(t_*,x_p(t_*))$: the periodic solution $x_p(t)$ and the constant solution $x_c$. That's a bit of a strange situation, because that means that if we take $(t_*,x_p(t_*))$ as the initial condition for our ODE $\dot{x} = f(x)$, that we can make a choice between two trajectories. In other words, this would imply non-uniqueness. If $f$ is locally Lipschitz, then by the Picard-Lindelöf theorem we know that this cannot happen, i.e. that $x_c$ and $x_p$ must be equal. So, if $f$ is locally Lipschitz, every periodic solution to $\dot{x} = f(x)$ is necessarily constant (and therefore not very exciting).
Interestingly enough, it turns out that you don't even need to assume that $f$ is locally Lipschitz to prove that nonconstant periodic solutions to $\dot{x} = f(x)$ do not exist. For more information, see @hardmath's answer to this question.
