My teacher proposed this "simple" proof that the 4-potential is a 4-vector which I am very skeptic about.
Since under gauge transformation the 4-four potential transforms as $$ A^\mu \mapsto A^\mu + \partial^\mu\lambda, $$ $\lambda$ being a scalar function, it follows that $A^\mu$ must trasform as a 4-vector under Lorentz transformation, since $\partial^\mu$ is one.
Is he right? What am I missing? I asked him for clarification but didn't get any more information other than this.
This works perfectly fine, but is more a heuristic.
If $\lambda$ is a scalar function, meaning that $\lambda \mapsto \lambda' = \lambda\circ \Lambda^{-1}$ under a Lorentz transformation $\Lambda$, then $\lambda(x) \mapsto \lambda'(x') = \lambda(\Lambda^{-1}\Lambda x) = \lambda(x)$. Most functions that look as if they are scalars are scalars, exceptions usually involve derivatives in some form. In particular $f(x) = x^0$ is a perfectly fine scalar function, even though it is not a Lorentz scalar in the sense that $f(\Lambda x) = f(x)$. It's confusing terminology.
In the same vein, transforming as a four-vector under Lorentz transformations means $A^\mu \mapsto \Lambda_\nu^\mu A'^\nu$ (note again the prime, standing for $A' = A\circ\Lambda^{-1}$), or $A\mapsto \Lambda\circ A\circ \Lambda^{-1}$, since then $A^\mu(x) \mapsto \Lambda^\mu_\nu A'^\nu(x') = \Lambda^\mu_\nu A^\nu(x)$ - the point at which you evaluate the function after the transformation still hasn't changed, but the transformation not only changed the way the coordinate is expressed (as $x'$ instead of $x$) but also the basis of your vector space.
So, finally, yes $\partial^\mu \lambda$ is a four-vector if $\lambda$ is a scalar function simply because $\partial^\mu \mapsto \Lambda^\mu_\nu \partial'^\nu$ and $\lambda(x)\mapsto \lambda'(x')$. Since adding two things that are not of the same type is generally not very well-defined, we conclude that $A^\mu$ better be a four-vector if it is to be a meaningful quantity. However, "better be" is not a proof. Formally you have to examine your definition of $A^\mu$ and deduce from that that it is a four-vector. How exactly that works depends on whether you cobbled it together from the non-relativistic parts $\phi,\vec A$ or defined it to be the anti-derivative of the field-strength tensor $F$.
Adding a four vector ($\partial^\mu \lambda$) to 4 components ($A^\mu$) does not necessarily mean the four components are a four vector. The usual proof that $A^\mu$ is a four vector follows from the wave equation $$\square\, A^\mu=4\pi j^\mu,$$ after using the covariance of the continuity equation to prove that $j^\mu$ is a four vector.
