I wouldn't call your argument a "proof" in the sense that you've assumed what you want to prove.
What could have happened is that the gravitational acceleration, $g$, was not universal but depended on the gravitational and inertial mass. To give a very simple model, suppose that
\begin{equation}
F = m_g g(m_g,m_i) = m_i g_{E} \left(\frac{m_g}{m_i}\right)^2
\end{equation}
where $g_E=9.8\ {\rm m/s}$ is the normal gravitational acceleration of Earth. Note in my notation, $g(m_g,m_i)$ is a function of the inertial mass $m_i$ and gravitational mass $m_g$ (with units of acceleration), while $g_E$ is just a number. On the right hand side, $g_E$ multiplies the square of the ratio of the gravitational and inertial masses (this is not true in Nature, it is just a silly model to prove a point).
Then using your argument, we would find
\begin{equation}
F = m_i a \implies a = g_E \left(\frac{m_g}{m_i}\right)^2
\end{equation}
In our hypothetical reality we could imagine that, say, feathers have $m_g/m_i=1$ and lead has $m_g/m_i=2$, then a pound of lead would fall four times faster than a pound of feathers (more precisely, an intertial-pound of lead would accelerate under the influence of gravity 4 times faster than an inertial-pound of features).
The non-trivial thing in gravity is that $g(m_g,m_i)=g_E$ exactly (as far as we can tell), for all objects. Fundamentally this is an experimental fact, and not something you can prove mathematically, but it has deep theoretical implications.
Your argument is not a proof because you assumed that $g(m_g,m_i)=g_E$ from the beginning, but this is precisely the thing you would need to prove, if you could prove inertial and gravitational mass are the same (but again, you can't prove this).