Suppose that $A$ is a finite abelian group. To make things more interesting, assume further that $A$ is an abelian $p$-group for some prime $p$.
Is it ever the case that $A$ can be written as $A = C \times D$, where $C$, $D$ are proper characteristic subgroups of $A$? (I am inserting the word "proper" so as to avoid having the trivial $C=1$, $D=A$.)
No. Both $C$ and $D$ are direct sums of cyclic $p$-groups, $$ \begin{align*} C &\cong C_{p^{a_1}}\oplus\cdots\oplus C_{p^{a_r}},\quad a_1\leq\cdots\leq a_r\\ D &\cong C_{p^{b_1}}\oplus\cdots \oplus C_{p^{b_t}},\quad b_1\leq\cdots\leq b_t. \end{align*} $$ If $a_i=b_j$ for any $i$ and $j$, then you have an automorphism of $G$ that swaps those two factors. This shows that neither $C$ nor $D$ are characteristic.
Otherwise, say $a_r\lt b_t$. Let the cyclic summands of $C$ be generated by $x_1,\ldots,x_r$, and the cyclic summands of $D$ be generated by $y_1,\ldots,y_t$. Define an automorphism of $G$ by fixing $x_1,\ldots,x_r$, $y_1,\ldots,y_{t-1}$, and mapping $y_t$ to $x_r+y_t$. The image of $D$ is not $D$, so $D$ is not characteristic.
To show $C$ is not characteristic either, fix $x_1,\ldots,x_{r-1}$, fix $y_1,\ldots,y_t$, and map $x_r$ to $x_r+p^{b_t-a_r}y_t$.
Of course, if you go to arbitrary finite abelian groups, you can decompose them into $p$-parts, each of which is characteristic.