Step 1: Reduction to Poisson equations. Take the divergence and curl of $\mathbf{F} = -\nabla\phi + \nabla\times\mathbf{A}$: $\nabla\cdot\mathbf{F} = -\nabla^2\phi$ (since $\nabla\cdot(\nabla\times\mathbf{A}) = 0$) and $\nabla\times\mathbf{F} = -\nabla^2\mathbf{A} + \nabla(\nabla\cdot\mathbf{A})$. Choosing the Coulomb gauge $\nabla\cdot\mathbf{A} = 0$: $\nabla\times\mathbf{F} = -\nabla^2\mathbf{A}$.

Step 2: Solve the Poisson equations. $\nabla^2\phi = -\nabla\cdot\mathbf{F}$ and $\nabla^2\mathbf{A} = -\nabla\times\mathbf{F}$. Using the Green's function $G(\mathbf{r}) = -1/(4\pi|\mathbf{r}|)$ for $\nabla^2$: $\phi = \frac{1}{4\pi}\int\frac{\nabla'\cdot\mathbf{F}}{|\mathbf{r}-\mathbf{r}'|}d^3r'$ and $\mathbf{A} = \frac{1}{4\pi}\int\frac{\nabla'\times\mathbf{F}}{|\mathbf{r}-\mathbf{r}'|}d^3r'$.

Step 3: Verify the decomposition. Define $\mathbf{F}_\ell = -\nabla\phi$ and $\mathbf{F}_t = \nabla\times\mathbf{A}$. Then $\nabla\times\mathbf{F}_\ell = 0$ (irrotational) and $\nabla\cdot\mathbf{F}_t = 0$ (solenoidal). We need $\mathbf{F}_\ell + \mathbf{F}_t = \mathbf{F}$.

Define $\mathbf{G} = \mathbf{F} - \mathbf{F}_\ell - \mathbf{F}_t$. Then $\nabla\cdot\mathbf{G} = \nabla\cdot\mathbf{F} + \nabla^2\phi - 0 = 0$ and $\nabla\times\mathbf{G} = \nabla\times\mathbf{F} - 0 + \nabla^2\mathbf{A} = 0$. A vector field with zero divergence and zero curl that decays at infinity is identically zero (it is both irrotational and solenoidal, hence harmonic and bounded, thus zero by Liouville's theorem for harmonic functions). Therefore $\mathbf{G} = 0$.

Step 4: Uniqueness. If $\mathbf{F} = -\nabla\phi_1 + \nabla\times\mathbf{A}_1 = -\nabla\phi_2 + \nabla\times\mathbf{A}_2$, then $\nabla(\phi_1-\phi_2) = \nabla\times(\mathbf{A}_1-\mathbf{A}_2)$. Taking divergence: $\nabla^2(\phi_1-\phi_2) = 0$ with decay at infinity, giving $\phi_1 = \phi_2$. Then $\nabla\times(\mathbf{A}_1-\mathbf{A}_2) = 0$ with gauge condition, giving $\mathbf{A}_1 = \mathbf{A}_2$. $\square$