From before, for X a continuous uniform variable, we get
\begin{gather*}
\int_u^v f(x) dx = \int_{u+\Delta}^{v+\Delta} f(x) dx\\
F(v)-F(u) = F(v+\Delta)-F(u+\Delta)\\
F(u+\Delta)-F(u) = F(v+\Delta)-F(v)\\
\frac{F(u+\Delta)-F(u)}{\Delta} = \frac{F(v+\Delta)-F(v)}{\Delta}
\end{gather*}
which is true regardless of \Delta so long as you stay in the domain of interest. Letting \(\Delta \rightarrow 0\) gives
\begin{equation*}
F'(u) = F'(v)
\end{equation*}
but since F is an antiderivative of the probability function,
\begin{equation*}
f(u) = f(v)
\end{equation*}
for all u and v in R. This only happens if f is constant...say, f(x)=c. If the space of X is a single interval with \(R = [a,b]\) then
\begin{equation*}
1 = \int_a^b c dx = c(b-a)
\end{equation*}
which yields \(c = \frac{1}{b-a}\) as desired.