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).