[proofplan]
We first verify that the proposed multiplication is a well-defined natural transformation $T^2 \Rightarrow T$. The two unit laws are exactly the two triangle identities for the adjunction, evaluated at the appropriate objects. The associativity law follows from the naturality of the counit $\varepsilon: FG \Rightarrow \operatorname{Id}_{\mathcal D}$ applied to the morphism $\varepsilon_{F c}: FGF c \to F c$, and then transported by the functor $G$.
[/proofplan]
[step:Define the multiplication as a natural transformation $T^2 \Rightarrow T$]
For every object $c \in \mathcal C$, define
\begin{align*}
\mu_c := G(\varepsilon_{F c}): GFGF c \to GF c.
\end{align*}
Since $\varepsilon: FG \Rightarrow \operatorname{Id}_{\mathcal D}$ is natural, the whiskered transformation
\begin{align*}
G\varepsilon F: GFGF \Rightarrow GF
\end{align*}
is natural. Hence $\mu := G\varepsilon F$ is a natural transformation $T^2 \Rightarrow T$, because $T^2 = GFGF$ and $T = GF$.
[/step]
[step:Verify the left unit law using the first triangle identity]
We prove
\begin{align*}
\mu \circ T\eta = \operatorname{id}_T.
\end{align*}
Let $c \in \mathcal C$ be an object. The component of $T\eta: T \Rightarrow T^2$ at $c$ is
\begin{align*}
(T\eta)_c = GF(\eta_c): GF c \to GFGF c.
\end{align*}
Therefore
\begin{align*}
(\mu \circ T\eta)_c
&= \mu_c \circ GF(\eta_c) \\
&= G(\varepsilon_{F c}) \circ G(F(\eta_c)) \\
&= G(\varepsilon_{F c} \circ F(\eta_c)).
\end{align*}
The triangle identity for the adjunction $F \dashv G$ gives
\begin{align*}
\varepsilon_{F c} \circ F(\eta_c) = \operatorname{id}_{F c}.
\end{align*}
Applying the functor $G$ gives
\begin{align*}
G(\varepsilon_{F c} \circ F(\eta_c))
= G(\operatorname{id}_{F c})
= \operatorname{id}_{GF c}.
\end{align*}
Thus $(\mu \circ T\eta)_c = \operatorname{id}_{T c}$ for every object $c \in \mathcal C$, so
\begin{align*}
\mu \circ T\eta = \operatorname{id}_T.
\end{align*}
[guided]
We need to prove the first monad unit law. Since natural transformations are equal exactly when their components are equal at every object, fix an object $c \in \mathcal C$.
The functor $T$ is $GF$, so the component of $T\eta$ at $c$ is obtained by applying $GF$ to the component $\eta_c: c \to GF c$:
\begin{align*}
(T\eta)_c = GF(\eta_c): GF c \to GFGF c.
\end{align*}
The multiplication component is
\begin{align*}
\mu_c = G(\varepsilon_{F c}): GFGF c \to GF c.
\end{align*}
Their composite is therefore
\begin{align*}
(\mu \circ T\eta)_c
&= \mu_c \circ GF(\eta_c) \\
&= G(\varepsilon_{F c}) \circ G(F(\eta_c)) \\
&= G(\varepsilon_{F c} \circ F(\eta_c)),
\end{align*}
where the last equality uses functoriality of $G$.
The first triangle identity for the adjunction says that the composite
\begin{align*}
F c \xrightarrow{F(\eta_c)} FGF c \xrightarrow{\varepsilon_{F c}} F c
\end{align*}
is the identity morphism on $F c$. Hence
\begin{align*}
\varepsilon_{F c} \circ F(\eta_c) = \operatorname{id}_{F c}.
\end{align*}
Applying $G$ yields
\begin{align*}
(\mu \circ T\eta)_c
= G(\operatorname{id}_{F c})
= \operatorname{id}_{GF c}
= \operatorname{id}_{T c}.
\end{align*}
Since this holds for every $c \in \mathcal C$, the natural transformations satisfy
\begin{align*}
\mu \circ T\eta = \operatorname{id}_T.
\end{align*}
[/guided]
[/step]
[step:Verify the right unit law using the second triangle identity]
We prove
\begin{align*}
\mu \circ \eta T = \operatorname{id}_T.
\end{align*}
Let $c \in \mathcal C$ be an object. The component of $\eta T: T \Rightarrow T^2$ at $c$ is
\begin{align*}
(\eta T)_c = \eta_{GF c}: GF c \to GFGF c.
\end{align*}
Thus
\begin{align*}
(\mu \circ \eta T)_c
&= \mu_c \circ \eta_{GF c} \\
&= G(\varepsilon_{F c}) \circ \eta_{GF c}.
\end{align*}
The second triangle identity for the adjunction, evaluated at the object $F c \in \mathcal D$, gives
\begin{align*}
G(\varepsilon_{F c}) \circ \eta_{GF c}
= \operatorname{id}_{GF c}.
\end{align*}
Therefore $(\mu \circ \eta T)_c = \operatorname{id}_{T c}$ for every object $c \in \mathcal C$, and hence
\begin{align*}
\mu \circ \eta T = \operatorname{id}_T.
\end{align*}
[/step]
[step:Prove associativity by applying naturality of the counit]
We prove
\begin{align*}
\mu \circ T\mu = \mu \circ \mu T.
\end{align*}
Let $c \in \mathcal C$ be an object. The two composites from $T^3 c = GFGFGF c$ to $T c = GF c$ are
\begin{align*}
(\mu \circ T\mu)_c
&= G(\varepsilon_{F c}) \circ GF(G(\varepsilon_{F c})), \\
(\mu \circ \mu T)_c
&= G(\varepsilon_{F c}) \circ G(\varepsilon_{FGF c}).
\end{align*}
It remains to prove these morphisms are equal.
Consider the morphism in $\mathcal D$
\begin{align*}
\varepsilon_{F c}: FGF c \to F c.
\end{align*}
Naturality of the counit $\varepsilon: FG \Rightarrow \operatorname{Id}_{\mathcal D}$ for this morphism gives the commutative square identity
\begin{align*}
\varepsilon_{F c} \circ FG(\varepsilon_{F c})
=
\varepsilon_{F c} \circ \varepsilon_{FGF c}.
\end{align*}
Applying the functor $G$ to this equality gives
\begin{align*}
G(\varepsilon_{F c}) \circ GFG(\varepsilon_{F c})
=
G(\varepsilon_{F c}) \circ G(\varepsilon_{FGF c}).
\end{align*}
Since $GFG(\varepsilon_{F c}) = GF(G(\varepsilon_{F c}))$, this is exactly
\begin{align*}
(\mu \circ T\mu)_c = (\mu \circ \mu T)_c.
\end{align*}
Because this holds for every object $c \in \mathcal C$, the natural transformations satisfy
\begin{align*}
\mu \circ T\mu = \mu \circ \mu T.
\end{align*}
[guided]
The associativity law says that the two ways of reducing
\begin{align*}
T^3 c = GFGFGF c
\end{align*}
to
\begin{align*}
T c = GF c
\end{align*}
must agree. Fix an object $c \in \mathcal C$. The first route applies $T\mu$ and then $\mu$:
\begin{align*}
(\mu \circ T\mu)_c
&= \mu_c \circ T(\mu_c) \\
&= G(\varepsilon_{F c}) \circ GF(G(\varepsilon_{F c})).
\end{align*}
The second route applies $\mu T$ and then $\mu$:
\begin{align*}
(\mu \circ \mu T)_c
&= \mu_c \circ \mu_{T c} \\
&= G(\varepsilon_{F c}) \circ G(\varepsilon_{F G F c}).
\end{align*}
We now identify the reason these two composites agree. The relevant theorem is the naturality of the counit $\varepsilon: FG \Rightarrow \operatorname{Id}_{\mathcal D}$. Apply it to the morphism
\begin{align*}
\varepsilon_{F c}: FGF c \to F c
\end{align*}
in $\mathcal D$. Naturality says that
\begin{align*}
\varepsilon_{F c} \circ FG(\varepsilon_{F c})
=
\varepsilon_{F c} \circ \varepsilon_{FGF c}.
\end{align*}
This equality takes place in $\mathcal D$ and compares the two ways of contracting adjacent $FG$ pairs before applying $G$.
Applying the functor $G$ to both sides gives
\begin{align*}
G(\varepsilon_{F c} \circ FG(\varepsilon_{F c}))
=
G(\varepsilon_{F c} \circ \varepsilon_{FGF c}).
\end{align*}
Using functoriality of $G$, this becomes
\begin{align*}
G(\varepsilon_{F c}) \circ GFG(\varepsilon_{F c})
=
G(\varepsilon_{F c}) \circ G(\varepsilon_{FGF c}).
\end{align*}
Finally, the expression $GFG(\varepsilon_{F c})$ is the same morphism as $GF(G(\varepsilon_{F c}))$, because both are obtained by applying the composite functor $GFG$ to $\varepsilon_{F c}$. Hence
\begin{align*}
G(\varepsilon_{F c}) \circ GF(G(\varepsilon_{F c}))
=
G(\varepsilon_{F c}) \circ G(\varepsilon_{FGF c}).
\end{align*}
This is precisely
\begin{align*}
(\mu \circ T\mu)_c = (\mu \circ \mu T)_c.
\end{align*}
Since the equality holds at every object $c \in \mathcal C$, we conclude
\begin{align*}
\mu \circ T\mu = \mu \circ \mu T.
\end{align*}
[/guided]
[/step]
[step:Conclude that the adjunction determines a monad]
The natural transformation $\eta: \operatorname{Id}_{\mathcal C} \Rightarrow T$ is the unit of the adjunction, and $\mu := G\varepsilon F: T^2 \Rightarrow T$ is a natural transformation. The two unit laws and the associativity law have been verified:
\begin{align*}
\mu \circ T\eta = \operatorname{id}_T,
\qquad
\mu \circ \eta T = \operatorname{id}_T,
\qquad
\mu \circ T\mu = \mu \circ \mu T.
\end{align*}
Therefore $(T, \eta, \mu)$ is a monad on $\mathcal C$.
[/step]
