I think it's true that $\varprojlim A_n\to\varinjlim A_n$ is always injective.

We may as well assume that

$A_1\leftarrowtail A_2\leftarrowtail A_3\leftarrowtail A_4\leftarrowtail \cdots$

is a sequence of inclusions of nested subgroups, so $\varprojlim A_n$ is just the intersection. An element of the kernel of $\varprojlim A_n\to\varinjlim A_n$ is just an element $a$ of $\bigcap A_n$ that is in the kernel of the map $A_1\twoheadrightarrow A_k$ for some $k$. But this implies $a=0$ since this map is a splitting of the inclusion $A_k\leftarrowtail A_1$.

This doesn't use countability.

I think it's true that $\varprojlim A_n\to\varinjlim A_n$ is always injective.

We may as well assume that

$A_1\leftarrowtail A_2\leftarrowtail A_3\leftarrowtail A_4\leftarrowtail \cdots$

is a sequence of inclusions of nested subgroups, so $\varprojlim A_n$ is just the intersection. An element of the kernel of $\varprojlim A_n\to\varinjlim A_n$ is just an element $a$ of $\bigcap A_n$ that is in the kernel of the map $A_1\twoheadrightarrow A_k$ for some $k$. But this implies $a=0$ since this map is a splitting of the inclusion $A_k\rightarrowtail A_1$.

This doesn't use countability.