[proofplan]
We verify the comonad axioms componentwise. The two counit identities follow directly from the two triangle identities for the adjunction $F \dashv G$. The coassociativity identity follows from naturality of the unit $\eta$ applied to the morphism $\eta_{G Y}: GY \to GFGY$, then applying the functor $F$.
[/proofplan]
[step:Define the comultiplication as a natural transformation]
Define the endofunctor
\begin{align*}
Q: \mathcal D &\to \mathcal D \\
Y &\mapsto F G Y
\end{align*}
on objects and by $Q(h) := F(Gh)$ on each morphism $h: Y \to Y'$ in $\mathcal D$.
Define
\begin{align*}
\delta: Q &\Rightarrow Q Q
\end{align*}
by setting, for each object $Y \in \mathcal D$,
\begin{align*}
\delta_Y := F(\eta_{G Y}): F G Y \to F G F G Y.
\end{align*}
Since $\eta: \operatorname{id}_{\mathcal C} \Rightarrow G F$ is natural, for every morphism $h: Y \to Y'$ in $\mathcal D$ the morphism $G h: GY \to GY'$ in $\mathcal C$ satisfies
\begin{align*}
G F(G h) \circ \eta_{G Y}
=
\eta_{G Y'} \circ G h.
\end{align*}
Applying the functor $F$ gives
\begin{align*}
F G F G h \circ F(\eta_{G Y})
=
F(\eta_{G Y'}) \circ F G h.
\end{align*}
This is precisely the naturality square for $\delta$, so $\delta: Q \Rightarrow Q Q$ is a natural transformation.
[/step]
[step:Verify the two counit identities from the triangle identities]
We must prove
\begin{align*}
\varepsilon_Q \circ \delta &= \operatorname{id}_Q, &
Q\varepsilon \circ \delta &= \operatorname{id}_Q.
\end{align*}
Let $Y \in \mathcal D$ be an object. The first triangle identity for the adjunction says that for every object $X \in \mathcal C$,
\begin{align*}
\varepsilon_{F X} \circ F(\eta_X) = \operatorname{id}_{F X}.
\end{align*}
Taking $X := GY$ gives
\begin{align*}
\varepsilon_{F G Y} \circ F(\eta_{G Y})
=
\operatorname{id}_{F G Y}.
\end{align*}
Since $(\varepsilon_Q)_Y = \varepsilon_{QY} = \varepsilon_{F G Y}$ and $\delta_Y = F(\eta_{G Y})$, this is
\begin{align*}
(\varepsilon_Q)_Y \circ \delta_Y
=
\operatorname{id}_{QY}.
\end{align*}
The second triangle identity says that for every object $Y \in \mathcal D$,
\begin{align*}
G(\varepsilon_Y) \circ \eta_{G Y}
=
\operatorname{id}_{G Y}.
\end{align*}
Applying $F$ gives
\begin{align*}
F G(\varepsilon_Y) \circ F(\eta_{G Y})
=
\operatorname{id}_{F G Y}.
\end{align*}
Since $(Q\varepsilon)_Y = Q(\varepsilon_Y) = F G(\varepsilon_Y)$, this becomes
\begin{align*}
(Q\varepsilon)_Y \circ \delta_Y
=
\operatorname{id}_{QY}.
\end{align*}
Thus both counit identities hold componentwise.
[/step]
[step:Prove coassociativity using naturality of the unit]
We must prove
\begin{align*}
\delta_Q \circ \delta = Q\delta \circ \delta.
\end{align*}
Let $Y \in \mathcal D$ be an object. Consider the morphism
\begin{align*}
\eta_{G Y}: GY \to G F GY
\end{align*}
in $\mathcal C$. Naturality of
\begin{align*}
\eta: \operatorname{id}_{\mathcal C} \Rightarrow G F
\end{align*}
applied to this morphism gives
\begin{align*}
G F(\eta_{G Y}) \circ \eta_{G Y}
=
\eta_{G F G Y} \circ \eta_{G Y}.
\end{align*}
Applying the functor $F$ to this equality yields
\begin{align*}
F G F(\eta_{G Y}) \circ F(\eta_{G Y})
=
F(\eta_{G F G Y}) \circ F(\eta_{G Y}).
\end{align*}
Now identify each side with the corresponding component of the comonad coassociativity diagram. Since
\begin{align*}
\delta_Y &= F(\eta_{G Y}), \\
(\delta_Q)_Y &= \delta_{QY} = F(\eta_{G F G Y}), \\
(Q\delta)_Y &= Q(\delta_Y) = F G(F(\eta_{G Y})) = F G F(\eta_{G Y}),
\end{align*}
the displayed equality is exactly
\begin{align*}
(Q\delta)_Y \circ \delta_Y
=
(\delta_Q)_Y \circ \delta_Y.
\end{align*}
Therefore
\begin{align*}
\delta_Q \circ \delta = Q\delta \circ \delta
\end{align*}
as natural transformations $Q \Rightarrow Q Q Q$.
[/step]
[step:Conclude that $F G$ is a comonad on $\mathcal D$]
The data
\begin{align*}
Q: \mathcal D &\to \mathcal D, &
\varepsilon: Q &\Rightarrow \operatorname{id}_{\mathcal D}, &
\delta: Q &\Rightarrow Q Q
\end{align*}
have been defined, and the two counit identities and the coassociativity identity have been verified. Hence $(Q, \varepsilon, \delta) = (F G, \varepsilon, F \eta G)$ is a comonad on $\mathcal D$.
[/step]
