As shown by Holevo, Shirokov, and Werner in (https://arxiv.org/abs/quant-ph/0504204), there exist separable but not countably decomposable states (in fact such states are dense as they conjecture). Since every separable state on the tensor product of finite dimensional Hilbert spaces is finitely decomposable, your condition 1 does not imply 2.

See the MO post Is the set of separable quantum states closed? for some relevant discussion (and my goof).

Your "countable separability" is called "countable decomposability" in [1]. As shown there, there exist separable but not countably decomposable states (in fact such states are dense as it is conjectured in [1]). Since every separable state on the tensor product of finite dimensional Hilbert spaces is finitely decomposable, your condition 1 does not imply 2.

See the MO post Is the set of separable quantum states closed? for some relevant discussion (and my goof).

[1] A. S. Holevo, M. E. Shirokov, R. F. Werner; Separability and Entanglement-Breaking in Infinite Dimensions. (https://arxiv.org/abs/quant-ph/0504204). See also (https://iopscience.iop.org/article/10.1070/RM2005v060n02ABEH000830).