I'm working on the problem 3.F(g) from Kelley' General Topology:
A real-valued function $f$ on a topological space $X$ is upper semi-continuous iff the set $\{x:f(x)\geq a\}$ is closed for each real number $a$. The upper topology $\mathscr{U}$ for the set $\mathbb R$ of real numbers consists of the void set, $\mathbb R$, and all sets of the form $\{t:t<a\}$ for $a$ in $\mathbb R$. Let $f$ be a non-negative real valued function on $X$, let $\mathbb R$ have the usual topology, and let $G=\{(x,t):0\leq t\leq f(x)\}$ have the relativized product topology of $X\times \mathbb R$. Let $\mathscr{D}$ be the decomposition of $G$ into "vertical slices"; that is, sets of the form $(\{X\}\times \mathbb R)\cap G$. If the decomposition $\mathscr{D}$ is upper semi-continuous, then $f$ is upper semi-continuous.
I have found a possible proof but I can't understand totally.
$\forall\ x\in X,\ \forall\ \varepsilon>0$, let $$V=\{y\in X:f(y)<f(x)+\varepsilon\},$$ $$\ H=\bigcup\ \left\{ \{z\}\times[0,f(z)]:\{z\}\times[0,f(z)]\subset\big{(}X\times[0,f(x)+\varepsilon)\big{)}\cap G \right\}.$$Then $P_{X}[H]=\{z\in X:f(z)<f(x)+\varepsilon\}=V$, where $P_{X}$ is the projection from $G$ to $X$. Notice that $\big{(}X\times[0,f(x)+\varepsilon)\big{)}\cap G$ is open in the space $G$ and it contains $H$. Since $\mathscr{D}$ is upper semi-continuous, the projection $P$ is closed, hence $H$ is open. Thus, $V$ is open for $P_{X}$ is open. So it is an open neighborhood of $x$. It follows that for each $x\in X$, $f$ is continuous relative to the upper topology $\mathscr{U}$, that is $f$ is upper semi-continuous.
Is this proof correct? I can't figure out why $\big{(}X\times[0,f(x)+\varepsilon)\big{)}\cap G$ is open in the space $G$. I'm also expect an alternative proof. Thanks for any help and hint!
Edit: A decomposition $\mathscr{D}$ of a topological space $X$ is upper semi-continuous iff for each $D$ in $\mathscr{D}$ and each open set $U$ containing $D$ there is an open set $V$ such that $D\subset V\subset U$ and $V$ is the union of members of $\mathscr{D}$.