[proofplan]
We compute the time derivative of the pulled-back two-form $\Omega_t:=\varphi_t^*\omega$.
\begin{align*}
\frac{d}{dt}\Omega_t=\varphi_t^*(\mathcal L_{X_t}\omega).
\end{align*}
Cartan's formula and the closedness of $\omega$ give
\begin{align*}
\mathcal L_{X_t}\omega=d(\iota_{X_t}\omega).
\end{align*}
Hence the condition $\frac{d}{dt}\Omega_t=0$ is equivalent to
\begin{align*}
\varphi_t^*d(\iota_{X_t}\omega)=0.
\end{align*}
Injectivity of pullback by the diffeomorphism $\varphi_t$ converts this into
\begin{align*}
d(\iota_{X_t}\omega)=0.
\end{align*}
Finally, a smooth family of forms with zero time derivative is constant in $t$, and the initial condition gives the required equality with $\omega$.
[/proofplan]
[step:Differentiate the pulled-back symplectic form along the isotopy]
For each $t\in[0,1]$, define the two-form $\Omega_t:=\varphi_t^*\omega\in\Omega^2(M)$. We use the standard pullback variation identity for a smooth family of diffeomorphisms with time-dependent generating vector field $X_t$, namely
\begin{align*}
\frac{d}{dt}\Omega_t=\frac{d}{dt}\varphi_t^*\omega=\varphi_t^*(\mathcal L_{X_t}\omega).
\end{align*}
Here
\begin{align*}
X_t=\frac{d\varphi_t}{dt}\circ\varphi_t^{-1}.
\end{align*}
This is precisely the vector field whose value at $y\in M$ is the velocity of the curve $t\mapsto\varphi_t(x)$ at the point $y=\varphi_t(x)$.
[guided]
We first isolate the object whose constancy is equivalent to being a symplectic isotopy. For each $t\in[0,1]$, set $\Omega_t:=\varphi_t^*\omega\in\Omega^2(M)$. The condition that $\varphi_t$ preserve the symplectic form is exactly the assertion that $\Omega_t=\omega$ for every $t$.
The relevant differentiation rule is the standard variation formula for pullbacks by a smooth one-parameter family of diffeomorphisms. It applies here because each $\varphi_t:M\to M$ is a diffeomorphism, the family is smooth in $t$, and the time-dependent vector field has been defined by $X_t=\frac{d\varphi_t}{dt}\circ\varphi_t^{-1}$. The formula gives $\frac{d}{dt}\varphi_t^*\omega=\varphi_t^*(\mathcal L_{X_t}\omega)$. Thus, instead of studying preservation of $\omega$ directly, we can study when the Lie derivative $\mathcal L_{X_t}\omega$ vanishes after pullback by $\varphi_t$.
[/guided]
[/step]
[step:Use Cartan's formula to reduce the derivative to an exterior derivative]
For each $t\in[0,1]$, define the one-form $\alpha_t:=\iota_{X_t}\omega\in\Omega^1(M)$. Cartan's formula for the Lie derivative gives
\begin{align*}
\mathcal L_{X_t}\omega=d(\iota_{X_t}\omega)+\iota_{X_t}d\omega.
\end{align*}
Since $(M,\omega)$ is symplectic, $d\omega=0$, so
\begin{align*}
\mathcal L_{X_t}\omega=d\alpha_t.
\end{align*}
Combining this with the pullback variation formula yields
\begin{align*}
\frac{d}{dt}\Omega_t=\varphi_t^*(d\alpha_t).
\end{align*}
[/step]
[step:Convert vanishing after pullback into closedness on $M$]
Fix $t\in[0,1]$. Since $\varphi_t:M\to M$ is a diffeomorphism, pullback by $\varphi_t$ is injective on differential forms:
\begin{align*}
\varphi_t^*\eta=0 \implies \eta=(\varphi_t^{-1})^*(\varphi_t^*\eta)=0.
\end{align*}
Applying this to the two-form $\eta=d\alpha_t$ gives
\begin{align*}
\frac{d}{dt}\Omega_t=0 \iff d\alpha_t=0.
\end{align*}
Since $\alpha_t=\iota_{X_t}\omega$, this is exactly the condition that $\iota_{X_t}\omega$ is closed.
[/step]
[step:Derive closedness from preservation of the symplectic form]
Assume first that $(\varphi_t)_{t\in[0,1]}$ is a symplectic isotopy. Then
\begin{align*}
\Omega_t=\varphi_t^*\omega=\omega
\end{align*}
for every $t\in[0,1]$. Hence
\begin{align*}
\frac{d}{dt}\Omega_t=0
\end{align*}
for every $t\in[0,1]$. By the equivalence established above, $d(\iota_{X_t}\omega)=0$ for every $t\in[0,1]$. Thus $\iota_{X_t}\omega$ is closed for every $t$.
[/step]
[step:Derive preservation of the symplectic form from closedness]
Conversely, assume that $\iota_{X_t}\omega$ is closed for every $t\in[0,1]$. Then
\begin{align*}
d\alpha_t=0
\end{align*}
for every $t$, where $\alpha_t=\iota_{X_t}\omega$. Therefore
\begin{align*}
\frac{d}{dt}\Omega_t=\varphi_t^*(d\alpha_t)=0
\end{align*}
for every $t\in[0,1]$.
It follows that the smooth family $\Omega_t$ is constant in $t$. Since $\varphi_0=\operatorname{id}_M$, we have
\begin{align*}
\Omega_0=\varphi_0^*\omega=\operatorname{id}_M^*\omega=\omega.
\end{align*}
Therefore $\Omega_t=\omega$ for every $t\in[0,1]$, that is, $\varphi_t^*\omega=\omega$. Thus $(\varphi_t)_{t\in[0,1]}$ is a symplectic isotopy.
[/step]
