I tried to prove the Part 1 as follows. Please check if it is valid.
We prove Part 1 by contradiction. Suppose the conclusion is not true, then $\max_{\overline{U}} u\neq \max_{\partial U} u.$ But since $\partial U\subset \overline{U},$ we have $\max_{\overline{U}} u>\max_{\partial U} u.$ Because $u\in C(\overline{U}),$ with $U$ bounded, it follows that $\overline{U}$ is compact in $\mathbb{R}^n,$ and so there exists $x_0\in \overline{U}$ such that $u(x_0)=\max_{\overline{U}}u.$ Since $\max_{\overline{U}} u>\max_{\partial U} u,$ we infer that $x_0\in U.$ In the case that $U$ is connected, by Part 2, we have $u\Big|_{\overline{U}}=u(x_0),$ which implies that $u\big|_{\partial{U}} =u(x_0),$ and so $\max_{\overline{U}} u=u(x_0)>\max_{\partial U} u=u(x_0),$ which is absurd. In the case that $U$ is disconnected, by the openness of $U,$ there is a non-empty connected component $W,$ such that $x_0$ is in the interior $\text{int }(W)$ of $W.$ Application of Part 2 to $u$ on $W,$ we get $u\Big|_{\overline{W}}=u(x_0),$ which leads to $u\Big|_{\partial W}=u(x_0),$ and so $\max_{\partial W} u=u(x_0).$ Since $\mathbb{R}^n$ is locally connected, we get that $\partial W\subset\partial U,$ and hence $\max_{\partial W}\leq \max_{\partial U}u,$ which implies that
$u(x_0)=\max_{\overline{U}} u>\max_{\partial U}u\geq \max_{\partial W}u=u(x_0),$ also a contradiction. Therefore, we have proved that $\max_{\overline{U}} u=\max_{\partial U} u.$