Consider a spacetime $(M,g)$ which admits a chart $(U, \varphi)$ with $\varphi^{-1}(p)=: x\in U$ for $p\in M$ and $\varphi^*g \equiv g_{\mu\nu}dx^{(\mu} \otimes dx^{\nu)}$ such that the spatial components of the metric $g_{ij}=g_{ij}(x^0)$ where $i,j \in \{1,2,3\}$, are non-trivial functions of time $x^0$. Such a region of spacetime $M$ as described by $(U, \varphi)$ is said to be dynamical.
This definition of whether a region of spacetime is dynamical needs to be checked for well-definedness. In other words, can there be another chart $(V,\psi)$ on $M$ such that the overlap $\varphi(U) \cap \psi(V)\neq \varnothing$ is stationary (i.e. non-dynamical)? If not, then it is well-defined.
I pointed to a well-known example in PhysSE where it seems to show exactly that this definition is not well-defined, that the notion of dynamics is coordinate-dependent.
However, there exists a coordinate-independent definition of a stationary spacetime, namely that it admits an asymptotically timelike Killing vector field.
How do we resolve this apparent paradox?
In your definition of dynamical you require only the existence of a (let's call it) a dynamical-chart. The existence of this chart doesn't rule out the possibility of a non-dynamical-chart but still it is a well defined property to possess a dynamical chart. For example consider the space-time $(\mathbb{R}^2, g)$ with $g = -x_2 dx_2\otimes dx_2 + dx_1\otimes dx_1 $, this is a non-dynamical chart but the spacetime is dynamical since it admits a dynamical chart, namely the chart induced by $\tilde{x} = x_1 + x_2$ and $\tilde{x_2} = x_2$, indeed $\frac {\partial}{\partial \tilde{x}_1 } = \frac {\partial}{\partial x_1 } - \frac {\partial}{\partial x_2 } $ and
$$g = -\tilde{x_2} d\tilde{x_2}\otimes d\tilde{x_2} + (1-\tilde{x_2})d\tilde{x_1}\otimes d\tilde{x_1} + \tilde{x_2} d\tilde{x_1}\otimes d\tilde{x_2} + \tilde{x_2} d\tilde{x_2}\otimes d\tilde{x_1}.$$
The existence of an asympt. timelike Killing vector guarantee the existence of a not-dynamical-chart.