Here's a counterexample to show that the image of a coalgebra map isn't always a coalgebra (adapted from the Nichols-Sweedler counterexamples).

Let $C = \mathbb{Z} \oplus \mathbb{Z}$ with $\Delta(e_1) = 0$, $\Delta(e_2)=e_1 \otimes e_1$ and $D = \mathbb{Z}/8 \oplus \mathbb{Z}/2$ with $\Delta(f_1) = 0$, $\Delta(f_2) = 4f_1 \otimes f_1$. (Here $e_i$ and $f_i$ denote the standard generators). Define a map $f\colon C \to D$ by $f(e_1) = 2f_1$ and $f(e_2) = f_2$. That's a coalgebra map.

Now the image of $f$ is $\mathbb{Z}/4 \oplus \mathbb{Z}/2$, I'm calling the generators $\overline e_1$ and $\overline e_2$. But it's not a sub-coalgebra. Any lift of $4f_1 = \Delta(f_2)$ to $im(f) \otimes im(f)$ would have to have order $4$, but $\overline e_2$ and $f_2$ only have order $2$.