The theorem is stated incorrectly.
In particular, consider the simple counter-example $u = -1$.
In this case, $Lu = -c \geq 0$ and clearly $u$ does not attain a nonnegative maximum.
Below is the correct statement with a very detailed proof that should answer all of your questions.
Theorem 5 (Weak Maximum Principle) Suppose that $\Omega$ is bounded and that $L$ is strictly elliptic with $c\leq0$.
If $u\in C^{2}(\Omega)\cup C(\overline{\Omega})$, $Lu\geq0$ in $\Omega$, and $\color{blue}{u\geq0 \text{ on } \partial \Omega}$, then a nonnegative maximum is attained at the boundary.
In the proof, we use the notation $a^{+}=\max\{a,0\}$.
Proof.
Let $\epsilon>0$ and $w(x)=\epsilon e^{\alpha x_{1}}+u(x)$.
Then,
\begin{multline*}
Lw(x)=\epsilon L[e^{\alpha x_{1}}](x)+Lu(x)\geq\epsilon L[e^{\alpha x_{1}}](x)\\
=\epsilon\left(\alpha^{2}a_{11}(x)+\alpha b_{1}(x)+c(x)\right)e^{\alpha x_{1}}\geq\epsilon\left(\alpha^{2}\lambda-\alpha\left\Vert b_{1}\right\Vert _{\infty}-\left\Vert c\right\Vert _{\infty}\right)e^{\alpha x_{1}}.
\end{multline*}
Since $\lambda>0$, we can always pick $\alpha$ large enough for $Lw>0$ to hold on all of $\Omega$.
Applying Lemma 4 of http://www.mi.uni-koeln.de/~gsweers/pdf/maxprinc.pdf, $w$ cannot attain a nonnegative maximum in $\Omega$.
As a result of this, $w^{+}$ cannot attain a maximum in $\Omega$ since then $w$ would attain a nonnegative maximum in $\Omega$.
We summarize this by writing $\sup_{\Omega}w^{+}=\sup_{\partial\Omega}w^{+}$.
Next, note that
$$
w^{+}=(u+\epsilon e^{\alpha x_{1}})^{+}\leq u^{+}+(\epsilon e^{\alpha x_{1}})^{+}=u^{+}+\epsilon e^{\alpha x_{1}}.
$$
Using the boundedness of $\Omega$, this implies that $\sup_{\partial\Omega}w^{+}\leq\sup_{\partial\Omega}u^{+}+\epsilon c$ for some large enough $c$.
Letting $\epsilon\rightarrow0$, we have $\sup_{\partial\Omega}w^{+}\leq\sup_{\partial\Omega}u^{+}$.
Lastly, since $u\leq w\leq w^{+}$, we can combine the inequalities in the previous two paragraphs to get $\sup_{\Omega}u\leq\sup_{\partial\Omega}u^{+}$.
Since $u\geq0$ on $\partial\Omega$, we have $\sup_{\partial\Omega}u^{+}=\sup_{\partial\Omega}u$.
Putting this all together,
$$
\sup_{\Omega}u\leq\sup_{\partial\Omega}u
$$
as desired. ∎
The other answer is probably better than this one; potential theory, polar sets, etc get to the heart of the matter. But for the disk specifically one can give a very elementary argument. Here "elementary" means "just using things in for example Rudin Real and Complex Analysis".
If $u$ is a bounded harmonic function in $\mathbb D$ then $f(e^{it})=\lim_{r\to1^-}u(re^{it})$ exists for almost every $t$. It's clear that $f$ is measurable, so $f\in L^\infty(\Bbb T)$. And in fact $u=P[f]$, the Poisson integral of $f$. Since the Poisson kernel is positive it's clear that $f\le0$ almost everywhere implies $u\le 0$.
(That's a proof for bounded harmonic functions in $\Bbb D$; I'm totally stumped on how to transfer it to a proof for bounded harmonic functions in $\frac12\Bbb D$, sorry.)
Best Answer
Suppose $a \leq b + \epsilon$ for all $\epsilon > 0$. Equivalently, $a < b + \epsilon$ for all $\epsilon > 0$. If $a>b$, then $a-b >0$, so $a < b + (a - b) = a$, contradiction. Hence, $a \leq b$.