To show 1), one can use the idea of "calibration". Given a smooth graphical subsolution to the minimal surface equation over say $B_1 \subset \mathbb{R}^n$, extend the upward unit normal $\nu$ to $B_1 \times \mathbb{R}$ by keeping it constant vertically. By the equation, $\text{div}(\nu) \leq 0$. We can use this to show that downward perturbations of the graph increase area. Indeed, for any domain $\Omega \subset B_1 \times \mathbb{R}$ that lies "beneath" the graph, the boundary consists of a part $\Sigma$ that lies on the graph, and a part $\Gamma$ that lies beneath. By the divergence theorem, $$0 \geq \int_{\Omega} \text{div} (\nu) = |\Sigma| + \int_{\Gamma} \nu \cdot w \geq |\Sigma| - |\Gamma|$$ where $w$ is the outer unit normal to $\Omega$. If a smooth graphical subsolution to the minimal surface equation touches a set of minimal perimeter from below somewhere, we can slide the graph up a little and apply this inequality to get a contradiction.
To show 2), use elementary geometry. Improvement of flatness says that for $\epsilon < \epsilon_0$ small, if a set of minimal perimeter $E$ satisfies $\partial E \cap \{|x| < 1\} \subset \{|x_{n+1}| < \epsilon\}$, then for some universal $\eta$, $\partial E \cap \{|\tilde{x}| < \eta\} \subset \{|\tilde{x}_{n+1}| < \epsilon \eta/2\}$ in a new coordinate system $(\tilde{x},\,\tilde{x}_{n+1})$. The "rotated" cylinder $\{|\tilde{x}| < \eta\} \cap \{|\tilde{x}_{n+1}| < \epsilon\eta/2\}$ (roughly) has to stay inside the original cylinder (otherwise we get points in $\partial E$ outside the original cylinder), so the angle between $e_{n+1}$ and $\tilde{e}_{n+1}$ is of order $\epsilon/\eta$. Iterating, we get that the normals rotate by order $2^{-k}\epsilon/\eta$ at the $k^{th}$ stage, and thus converge geometrically.