The answer is:that's not true.
it is look like
- generate $(X_1,X_2,X_3)$ such that $S=X_1+X_2+X_3 \leq 1$
- $X_4=(1-S)$
or on the other hands $X_4= (1-S|S<1) $.
now we want to find a conditional distribution $(1-S|S<1)$.
$S=X_1+X_2+X_3$ and according to https://en.wikipedia.org/wiki/Irwin%E2%80%93Hall_distribution
\begin{eqnarray}
f_S(a)=\left\{
\begin{array}{cc}
\frac{1}{2} a^2 & 0\leq a <1 \\
\frac{1}{2}(-2a^2+6a-3) & 1\leq a<2 \\
\frac{1}{2} (a-3)^2 & 2\leq a<3
\end{array}
\right.
\end{eqnarray}
so $P(S\leq 1)=\int_{0}^{1} \frac{1}{2} a^2 da=\frac{1}{6}$
and
$P( 1-x \leq S\leq 1)=\int_{1-x}^{1} \frac{1}{2} a^2 da$
$=\frac{1}{6} a^3|_{1-x}^{1}=\frac{1}{6}(1-(1-x)^3)$
so $F_{X_4}(x)=P(X_4\leq x)=P(1-S\leq x|S<1)=P(S\geq 1-x|S\leq 1)$
$=\frac{P( 1-x \leq S\leq 1)}{P(S\leq 1)}=1-(1-x)^3 $ $,x<1$
its not look like uniform distribution and $f_{X_4}(x)=3(1-x)^2$ .($beta(1,3)$ distribution)
a simple simulation confirms that the $X_4$ does not follow uniform distribution:
x4<-c()
count<-1
simu<-6*1000
for(i in 1:simu){
x<-runif(3)
s<-sum(x)
if(s<1) {x4[count]<-1-s;count<-count+1}
}
plot(density(x4),type="l",xlim=c(0,1))
> ks.test(x4,runif(length(x4)))
Two-sample Kolmogorov-Smirnov test
data: x4 and runif(length(x4))
D = 0.37026, p-value < 2.2e-16
alternative hypothesis: two-sided

it's simple to check $1-S|S>1$ does not have uniform distribution.