Doubt in the demonstration of the Half Space Theorem ($\textbf{half-space Theorem}$) A connected minimal surface $S$,
without boundary, properly immersed can not be contained in a half-space,
unless $S$ is a plane.
Let $C_{c}$ the catenoid.
Consider the set $Q_{c}=\left\{ \left( x_{1},x_{2},x_{3}\right) \in 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
^{3};\text{ \ existem }\left( x_{1},x_{2},\widetilde{q}_{3}\right) \in
C_{c}^{-}\text{ e }\left( x_{1},x_{2},q_{3}\right) \in S\text{ com }%
\widetilde{q}_{3}\leq x_{3}\leq q_{3}\right\} $, where $c\in \left[ 0,R%
\right] $ with $R>0$ fixe.
Let $L:=\left\{ r\in \left( 0,R\right) ;Q_{r}\neq \varnothing \right\} $ and
suposed that ther $\sup L:=\widetilde{c}$.
Show that $S\geq C_{\widetilde{c}}$ in the third coordinate.
I am reading the demonstration of this famous theorem. But I was a little confused when the author uses this function h and mixes with the sup and says that it is continuous and so on. It does not look wrong, but it's strange. Could someone explain me?
To help, I attached a proof print. I apologize for being in Portuguese, but I did not get an English version, but you can understand.

 A: The point of defining this auxiliary function $h$ is that $h(c) \geq 0$ precisely if $C_{c}^{-}$ and $S$ intersect on the interior. 
You can see from construction in the reference you gave that $C_{c}^{-}  \to \Pi_{t_{1}}$ smoothly (away from the origin) as $c \to 0$ for some $t_{1} > t_{0}$, and since $S$ asymptotically approaches the plane $\Pi_{t_{0}}$ we know that for small enough $c$ that $C_{c}^{-} \cap S \neq \emptyset$, which is equivalent to $Q_{c} \neq \emptyset$.
The comments about the well-definedness and continuity of $h$ are just a basic exercise in real analysis. $C_{c}^{-} \cap \Pi^{+}$ is a compact portion of the surface bounded by a circle of radius $c$ and a circle in the plane $\Pi_{t_{0}}$, and $S$ is closed as a subset of $\Pi^{+}$, you can show by taking a maximizing sequence that the supremum of the set
$$ \{ \tilde{q}_{3} - q_{3} : (x_{1}, x_{2}, \tilde{q}_{3}) \in C_{c}^{-} \cap \Pi^{+}, (x_{1}, x_{2}, q_{3}) \in S\}$$
is achieved, which is to say $h$ is well-defined as a function in $c$. Continuity then is just a simple consequence of the fact that $\tilde{q}_{3}$ is continuous as a function in $c$ (You can check this just from writing down the parametrization of the catenoid). 
Now the point is that if $\tilde{c} = \sup\{c \in (0, R): Q_{c} \neq \emptyset\}$ (by construction $Q_{R} = \emptyset$ so we know this supremum exists and is $< R$) then we have to show that $h(\tilde{c}) = 0$ (the zeroes of $h$ are precisely given by the radii where $S \cap (C_{c}^{-} \cap \Pi^{+}) \neq \emptyset$ and $S$ remains on one side of $C_{c}^{-}$, which are the conditions for the geometric interior maximum principle), which implies that $S$ and $C_{\tilde{c}}^{-}$ touch on the interior (since we know $S$ doesn't intersect $C_{c}^{-}$ near the boundary circle of radius $c$ by construction), and so the maximum principle for minimal surfaces applies, so $S$ is a portion of the catenoid, which contradicts the fact that $S \subset \Pi^{+}$ since $S$ is closed and therefore equal $C_{c}$, which is never contained in a half space. So to do this the author applies continuity of $h$ while assuming $h(\tilde{c}) > 0$ to see there is an $\epsilon > 0$ so that $h(\tilde{c} + \epsilon) > 0$ which implies $Q_{\tilde{c} + \epsilon} \neq \emptyset$, so we have a contradiction and must have $h(\tilde{c}) = 0$ as desired.
