let $\Omega$ be an open, connected and bounded subset of $\mathbb{R}^\mathbb{N}$ and $u \in C^2(\Omega)\cap C(\overline{\Omega})$.
In most books in order to prove that if $u$ is harmonic in $\Omega$ then: $$\max_{\overline{\Omega}} u = \max_{\partial \Omega} u,$$ they usually assume a function$$u_ε =u +ε |x|^2, \quad ε \geq 0,$$ and then suppose that $u_ε$ attains its maximum at an interior point $\overline{x} \in \Omega$, then we stumble into a contradiction since $\Delta u_ε = 2nε \geq 0$ is the trace of the hessian matrix which is negative definite at $\overline{x}$ and finally we conclude by letting $ε \to 0^+$
My question is what difference does it make if we work with $u$ alone directly? We have that $\Delta u = 0$ and the Hessian matrix is negative definite as well, meaning no $0$ eigenvalue. Isn't that a contradition as well?
You don't get a contradiction if you deal with $u$ alone; you only get that the Hessian is negative semi-definite, meaning it could still have $0$ as an eigenvalue. The contradiction comes from a strict inequality. Here is the full argument:
Suppose that $u \in C(\overline \Omega) \cap C^2(\Omega)$ is harmonic in $\Omega$ (where $\Omega$ is open, bounded and connected). We would like to say that if $u$ attains a maximum in $\Omega$ then $\Delta u < 0$ which contradicts that $u$ is harmonic. However, if $u$ attains a maximum in $\Omega$ then we only know that $\Delta u \le 0$ which does not give a contradiction. Instead, we consider $u^\epsilon = u + \epsilon \lvert x \rvert^2$ for $\epsilon > 0$. Now if $u^\epsilon$ attains a maximum in $\Omega$, then $\Delta u^\epsilon \le 0$. However, this is impossible because $\Delta u^\epsilon = 2n\epsilon > 0$ (note the strict inequality). Thus the maximum of $u^\epsilon$ must occur on the boundary. Now $u \le u^\epsilon$, so we see $$\max_{\overline \Omega} u \le \max_{\overline \Omega} u^\epsilon = \max_{\partial \Omega} u^\epsilon \le \max_{\partial \Omega} u + \epsilon R^2$$ where $\Omega$ is contained in the ball of radius $R$ centered at the origin. Now taking $\epsilon \searrow 0$, we see $$\max_{\overline \Omega} u \le \max_{\partial \Omega} u$$ as desired.