Note, that there is a function $h \colon \bar B \to \mathbf R$, which is harmonic on $\bar B$ and has $h|_{\partial B} = u$. As $u$ is subharmonic, this implies $u \le h$. Hence,
$$ u(x) \le h(x) \le \sup_{\partial B} h = \sup_{\partial B} u = u(x) $$
That is $h(x) = \sup_{\partial B} h$. As $h$ is harmonic on $B$, by the strong maximum principle for harmonic function (which you already proved, I think), we have, that $h$ is constant on $\bar B$, hence, $u$ is constant on $\partial B$, as $h = u$ on $\partial B$.
If $u$ has a local maximum at $x_0 \in \Omega$, that means that there is an open neighbourhood $U\subset \Omega$ of $x_0$ such that $u\lvert_U$ has a global maximum at $x_0$. Since clearly $u\lvert_U$ is harmonic on $U$, by the form of the maximum principle that you already know, it follows that $u\lvert_U$ is constant.
The two harmonic functions $u$ and $v \colon z \mapsto u(x_0)$ hence coincide on a nonempty open set (namely on $U$), and since $\Omega$ is connected, and harmonic functions are real-analytic, the following identity theorem (the case $n = 2$ with the usual identification $\mathbb{C}\cong \mathbb{R}^2$) implies that $u \equiv v$ on $\Omega$, i.e. $u$ is constant.
Identity theorem: Let $W\subset \mathbb{R}^n$ a connected open set, and $f,g \colon W \to \mathbb{R}$ real-analytic. If there is a nonempty open set $V\subset W$ such that $f\lvert_V \equiv g\lvert_V$, then $f \equiv g$ on all of $W$.
Proof: It suffices to consider $g \equiv 0$, since $f\equiv g \iff f-g \equiv 0$. Consider the set
$$Z = \{ x \in W : \text{there is a neighbourhood } U \text{ of } x \text{ such that } f\lvert_U \equiv 0\}.$$
By its definition, $Z$ is an open subset of $W$ (if $x\in W$ and $U$ is a neighbourhood of $x$ such that $f\lvert_U \equiv 0$, then $y\in Z$ for all $y$ in the interior of $U$). Since $f$ is real analytic, we can also describe $Z$ as
$$Z = \{ x \in W : D^{\alpha}f(x) = 0 \text{ for all } \alpha \in \mathbb{N}^n\},$$
for the power series representation of $f$ about $x$,
$$f(y) = \sum_{\alpha \in \mathbb{N}^n} \frac{D^{\alpha}f(x)}{\alpha!}(y-x)^{\alpha},$$
vanishes in a neighbourhood of $x$ if and only if all coefficients vanish. By continuity of $D^{\alpha}f$, the set
$$Z_{\alpha} = (D^{\alpha}f)^{-1}(0)$$
is closed (in $W$) for every $\alpha \in \mathbb{N}^n$, and hence so is the intersection
$$Z = \bigcap_{\alpha \in \mathbb{N}^n} Z_{\alpha}.$$
Thus $Z$ is an open and closed subset of $W$, and since $W$ is connected, we have either $Z = W$ or $Z = \varnothing$. But $Z \neq \varnothing$ was part of the hypotheses of the theorem, so $Z = W$ and indeed $f \equiv 0$ on $W$.
Best Answer
$\newcommand{\R}{\mathbb{R}}$ $\newcommand{\vp}{\varphi}$
If I understand correctly, this is a consequence of the standard maximum principle.
If $q \colon \R^2 \to \R$ is a linear functional, then the composition $q \circ \vp \colon \overline{B} \to \R$ is a harmonic function, and we can apply the standard maximum principle. More precisely, if $C \subseteq \R^2$ is closed and convex, and $\vp(\partial B) \subseteq C$, then the maximum principle tells us $$ q \circ \vp \le \sup_{x \in \partial B} q(\vp(x)) \le \sup_{y \in C} q(y) \quad \text{on } \overline{B}. $$ Since the set $C$ is closed and convex, it can be described as an intersection of a family (possibly infinite) of half-planes. More precisely, $$ C = \bigcap_{q \colon \R^2 \to \R \text{ linear}} \big\{ y \in \R^2 : q(y) \le \sup_{y \in C} q(y) \big\}. $$ By the previous reasoning, each point $\vp(x)$ (for $x \in \overline{B}$) lies in every set of the form $\big\{ y \in \R^2 : q(y) \le \sup_{y \in C} q(y) \big\}$, and so we showed that $\vp(x) \in C$ for all $x \in \overline{B}$.
With some care, one can apply the strong maximum principle for $q \circ \vp$ and infer a stronger statement for $\vp$ (that is, $\vp(x) \in \Omega$ for $x \in B$, not just $\vp(x) \in \overline{\Omega}$).