An elementary direct strategy to prove the result is the following. Let us make things simpler, at least initially, referring to the elementary equation in $\mathbb R$ (the procedure easily extends to $\mathbb R^n$ and to Hamilton equations in particular)

$$\frac{dx}{dt} = F(x(t))\:.$$

Let us assume that two solutions $x$ and $x'$ exist with the same initial condition $x(0)=x_0= x'(0)$. Finally suppose that $dF/dx$ is bounded by some positive real $L<+\infty$ (if $dF/dx$ is continuous, it is automatically true working in a bounded neighbourhood).

I go to prove that it must be $x(t)=x'(t)$ in a sufficiently small right neighbourhood of $t=0$.

The considered differential equation is equivalent to the integral one (as Ron said)

$$x(t) = x_0 + \int_0^t F(x(u)) du\:,$$

so that, we also have

$$x'(t) = x_0 + \int_0^t F(x'(u)) du\:,$$

and thus

$$x(t)-x'(t) = \int_0^t F(x(u)) -F(x'(u)) du \:.$$

Therefore, using Lagrange theorem (below, $y(u)$ is some unknown point between $x(u)$ and $x'(u)$),

$$|x(t)-x'(t)| \leq \int_0^t |F(x(u)) -F(x'(u))| du = \int_0^t \left|\frac{dF}{dx}|_{y(u)}(x'(u)-x(u)) \right|du$$

We can now use the bound $|dF/dx| \leq L$, obtaining

$$|x(t)-x'(t)| \leq L \int_0^t |x(u)-x'(u)| du \leq Lt \max_{u \in [0,t]} |x(u)-x'(u)|$$

Obviously, if $0\leq t\leq T$

$$Lt \max_{u \in [0,t]} |x(u)-x'(u)| \leq LT \max_{u \in [0,T]} |x(u)-x'(u)|$$

so that

$$|x(t)-x'(t)| \leq LT \max_{u \in [0,T]} |x(u)-x'(u)| \quad \forall t\in [0,T]$$

which eventually produces, since the bound must hold also for the value $t=t_0$ where the (continuous) function in the left-hand side above attains its maximum,

$$ \max_{t \in [0,T]} |x(t)-x'(t)| \leq LT \max_{u \in [0,T]} |x(u)-x'(u)|\:, $$

i.e.

$$0 \leq \max_{t \in [0,T]} |x(t)-x'(t)| \leq LT \max_{t \in [0,T]} |x(t)-x'(t)|\:. $$

We can take $T>0$ so small that $0< LT<1$. In this case the found inequality can hold only if

$$ \max_{t \in [0,T]} |x(t)-x'(t)| =0\:,$$

and thus $x(t)=x'(t)$ for $t\in [0,T]$.

We have found that, at least in a sufficiently small neighbourhood $[0,T]$ of the initial time (however the procedure can be implemented also for $t<0$), $x(t)=x'(t)$. This reasoning, gluing together several patches, establishes the uniqueness of the solution on its whole domain.

Let us pass to the true Hamilton equations. The proof is essentially identical for a $C^2$ Hamiltonian $H$ defined an open set of $\mathbb R^{2n}$, with the following replacements.

$${\mathbb R} \to {\mathbb R}^{2n}$$

$$x \to \vec{x} = (\vec{q}, \vec{p})$$

$$ F \to \vec{F}(\vec{x}) := S\nabla_{\vec{x}} H(\vec{x})$$

where $S$ is the symplectic $2n \times 2n $ matrix with $-I$, $I$ on the anti diagonal and $0$, $0$ on the diagonal when viewed as 2x2 block matrix.

$$ |\cdot | \to ||\cdot|| $$

the Euclidean norm in $\mathbb R^{2n}$.