To see that every countryman line is Aronszajn line, it suffices to show that $\omega_1$, $\omega_1^*=(\omega_1,\ni)$ and uncountable separable liner order are not a Countryman line. (This is because that the subset of a Countryman line is also a Countryman line.)
I cannot prove that every uncountable separable linear order is not a Countryman line. How to prove this.
The proof of that $\omega_1$ is not a Countryman line is :
Assume that there are countably many linear orders $C_n\subset\omega_1\times\omega_1$ with $\cup\{C_n|n\in\omega_1\}=\omega_1\times\omega_1$. For every $\alpha\in\omega_1$ choose $n_\alpha$ such as $D_\alpha=\{y\in\omega_1|(x,y)\in C_{n_\alpha}\}$ is uncounable. Then there are $\alpha\in\beta\in\omega_1$ such that $n_\alpha=n_\beta$ (by the Pigeonhole principle). Fix arbitrary $\delta\in\omega_1$ such as $(\beta,\delta)\in C_{n_\alpha}$. Then since $C_{n_\alpha}$ is uncountable, there is $\gamma$ such that $\delta\in\gamma$ and $(\alpha,\gamma)\in C_{n_\alpha}$. This shows that there is incompatible pair $(\alpha,\gamma)$ and $(\beta,\delta)$ in $C_{n_\alpha}=C_{n_\beta}$, a contradiction.
Note. Following are definition of Aronszajn line and Countryman line:
Aronszajn line is an uncountable line which does not possess a sublinear order which isomorphic to $\omega_1$, $\omega_1^*$ or an uncountable linear order.
Countryman line ,(L,<_L), is a line such that there are countably many sublinear orders $C_n\subset L\times L$ such that $\cup\{C_n|n\in\omega\}=L\times L$. Where the order on $L\times L$ is given by $(a,b)<_L(c,d)$ provided that "$a\leq_Lc$ and $b\leq_Ld$" and "$\neg(a=c\land b=d)$".