[proofplan]
The proof is an unpacking of the definition of a Hamiltonian isotopy under the sign convention used here. In the forward direction, a Hamiltonian isotopy is generated by Hamiltonian vector fields, so its generator satisfies the exactness equation by definition. In the reverse direction, the nondegeneracy of $\omega$ makes the assignment $Y\mapsto\iota_Y\omega$ injective on vector fields, so any generator whose contraction is $dH_t$ must be the Hamiltonian vector field of $H_t$.
[/proofplan]
[step:Use the definition of a Hamiltonian isotopy to obtain exactness]
Assume first that $(\varphi_t)_{t\in[0,1]}$ is Hamiltonian. By definition, there exists a smooth map $H:[0,1]\times M\to\mathbb R$ such that, writing $H_t:M\to\mathbb R$ for the smooth function $x\mapsto H(t,x)$, the isotopy is generated by the time-dependent Hamiltonian vector field $X_{H_t}\in\mathfrak X(M)$.
Thus $\varphi_t$ satisfies the differential equation
\begin{align*}
\frac{d\varphi_t}{dt}=X_{H_t}\circ\varphi_t.
\end{align*}
By the definition of the generating vector field of the isotopy,
\begin{align*}
X_t=\frac{d\varphi_t}{dt}\circ\varphi_t^{-1}.
\end{align*}
Composing the Hamiltonian flow equation on the right with $\varphi_t^{-1}$ gives
\begin{align*}
X_t=X_{H_t}.
\end{align*}
Using the stated sign convention for Hamiltonian vector fields,
\begin{align*}
\iota_{X_t}\omega=\iota_{X_{H_t}}\omega=dH_t.
\end{align*}
Therefore $\iota_{X_t}\omega$ is exact for every $t\in[0,1]$.
[/step]
[step:Use nondegeneracy of $\omega$ to recover the Hamiltonian vector field]
Conversely, suppose there exists a smooth map $H:[0,1]\times M\to\mathbb R$ such that, for $H_t(x)=H(t,x)$,
\begin{align*}
\iota_{X_t}\omega=dH_t
\end{align*}
for every $t\in[0,1]$.
For each $t\in[0,1]$, define $Y_t\in\mathfrak X(M)$ to be the Hamiltonian vector field of $H_t$ with the sign convention of the statement, so
\begin{align*}
\iota_{Y_t}\omega=dH_t.
\end{align*}
This vector field is unique: for each $x\in M$, the [linear map](/page/Linear%20Map)
\begin{align*}
T_xM&\to T_x^*M,
\end{align*}
\begin{align*}
v&\mapsto \iota_v\omega_x
\end{align*}
is an isomorphism because $\omega_x$ is nondegenerate.
Since both $X_t$ and $Y_t$ satisfy the same contraction identity,
\begin{align*}
\iota_{X_t}\omega=dH_t=\iota_{Y_t}\omega,
\end{align*}
we have
\begin{align*}
\iota_{X_t-Y_t}\omega=0.
\end{align*}
Pointwise nondegeneracy of $\omega$ implies $X_t-Y_t=0$, hence
\begin{align*}
X_t=Y_t=X_{H_t}.
\end{align*}
[guided]
We now prove the reverse implication with the uniqueness mechanism made explicit. Suppose that a smooth map $H:[0,1]\times M\to\mathbb R$ is given, and write $H_t:M\to\mathbb R$ for the smooth function $x\mapsto H(t,x)$. The hypothesis says that the generating vector field $X_t$ of the isotopy satisfies
\begin{align*}
\iota_{X_t}\omega=dH_t
\end{align*}
for every $t\in[0,1]$.
The definition of a Hamiltonian vector field under the sign convention used here is precisely this: the Hamiltonian vector field of $H_t$ is the unique vector field $Y_t\in\mathfrak X(M)$ satisfying
\begin{align*}
\iota_{Y_t}\omega=dH_t.
\end{align*}
Why is there a unique such vector field? At each point $x\in M$, nondegeneracy of the symplectic form means that the linear map
\begin{align*}
T_xM&\to T_x^*M,
\end{align*}
\begin{align*}
v&\mapsto \iota_v\omega_x
\end{align*}
has zero kernel and, since $T_xM$ and $T_x^*M$ have the same finite dimension, is an isomorphism. Therefore a one-form at $x$ determines a unique tangent vector at $x$.
Apply this to the one-form $(dH_t)_x$. The vector $(X_t)_x$ satisfies
\begin{align*}
\iota_{(X_t)_x}\omega_x=(dH_t)_x,
\end{align*}
and the vector $(Y_t)_x$ satisfies
\begin{align*}
\iota_{(Y_t)_x}\omega_x=(dH_t)_x.
\end{align*}
Subtracting these two equalities gives
\begin{align*}
\iota_{(X_t)_x-(Y_t)_x}\omega_x=0.
\end{align*}
Because the contraction map is injective, this implies
\begin{align*}
(X_t)_x=(Y_t)_x.
\end{align*}
Since this holds for every $x\in M$, we obtain $X_t=Y_t=X_{H_t}$ as vector fields on $M$.
[/guided]
[/step]
[step:Conclude that the isotopy solves the Hamiltonian flow equation]
From the previous step, $X_t=X_{H_t}$ for every $t\in[0,1]$. Since $X_t$ is the generating vector field of $(\varphi_t)$,
\begin{align*}
\frac{d\varphi_t}{dt}=X_t\circ\varphi_t=X_{H_t}\circ\varphi_t.
\end{align*}
Thus $(\varphi_t)$ is the flow of the time-dependent Hamiltonian vector field associated to $H_t$. By definition, $(\varphi_t)_{t\in[0,1]}$ is a Hamiltonian isotopy.
[/step]
