[proofplan]
We argue by contradiction from the definition of a [discrete random variable](/page/Discrete%20Random%20Variable). If $U$ were discrete, then there would be a [countable set](/page/Countable%20Set) $S\subset\mathbb R$ such that $\mathbb P(U\in S)=1$. The continuous uniform law assigns probability $0$ to every singleton, and countable additivity then forces every countable set to have probability $0$, contradicting the assumed full-probability support.
[/proofplan]
[step:Record that every singleton has zero probability under the continuous uniform law]
Let $\mu_U:=\mathbb P\circ U^{-1}$ denote the law of $U$, so $\mu_U$ is a probability measure on $(\mathbb R,\mathcal B(\mathbb R))$. Since $U\sim\operatorname{Unif}(0,1)$, for every Borel set $B\in\mathcal B(\mathbb R)$ we have
\begin{align*}
\mu_U(B)=\mathcal L^1(B\cap(0,1)).
\end{align*}
Fix $a\in\mathbb R$. The singleton $\{a\}$ belongs to $\mathcal B(\mathbb R)$, and therefore the event $\{U=a\}=U^{-1}(\{a\})$ belongs to $\mathcal F$. Using the defining formula for the continuous uniform law,
\begin{align*}
\mathbb P(U=a)=\mu_U(\{a\})=\mathcal L^1(\{a\}\cap(0,1))=0.
\end{align*}
Thus every singleton value has probability $0$.
[guided]
We first translate the hypothesis about the continuous uniform distribution into a statement about individual points. Let $\mu_U:=\mathbb P\circ U^{-1}$ denote the law of $U$, so $\mu_U$ is a probability measure on the measurable space $(\mathbb R,\mathcal B(\mathbb R))$. By the stated meaning of the continuous uniform distribution on $(0,1)$, for every Borel set $B\in\mathcal B(\mathbb R)$ we have
\begin{align*}
\mu_U(B)=\mathcal L^1(B\cap(0,1)).
\end{align*}
Fix $a\in\mathbb R$. The singleton $\{a\}$ is a Borel subset of $\mathbb R$, so $U^{-1}(\{a\})\in\mathcal F$ because $U$ is measurable. This event is exactly $\{U=a\}$. Applying the law formula with $B=\{a\}$ gives
\begin{align*}
\mathbb P(U=a)=\mu_U(\{a\})=\mathcal L^1(\{a\}\cap(0,1)).
\end{align*}
The set $\{a\}\cap(0,1)$ is either empty or a singleton. In either case its one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) is $0$, and hence
\begin{align*}
\mathbb P(U=a)=0.
\end{align*}
Thus every singleton value has probability $0$ under the law of $U$.
[/guided]
[/step]
[step:Contradict the existence of a countable support of full probability]
Assume, toward a contradiction, that $U$ is discrete. By the definition of a discrete [random variable](/page/Random%20Variable), there exists a countable set $S\subset\mathbb R$ such that
\begin{align*}
\mathbb P(U\in S)=1.
\end{align*}
If $S=\varnothing$, then $\{U\in S\}=\varnothing$ and $\mathbb P(U\in S)=0$, contradicting $\mathbb P(U\in S)=1$. Hence $S\ne\varnothing$.
Since $S$ is countable and nonempty, choose an index set $I$ equal either to $\{1,\dots,N\}$ for some $N\in\mathbb N$ or to $\mathbb N$, and choose an injective enumeration $(s_m)_{m\in I}$ of $S$. For each $m\in I$, define the event
\begin{align*}
A_m:=\{\omega\in\Omega:U(\omega)=s_m\}.
\end{align*}
The events $(A_m)_{m\in I}$ are pairwise disjoint, and
\begin{align*}
\{U\in S\}=\bigcup_{m\in I}A_m.
\end{align*}
By finite additivity if $I$ is finite, and by countable additivity if $I=\mathbb N$,
\begin{align*}
\mathbb P(U\in S)=\sum_{m\in I}\mathbb P(A_m).
\end{align*}
The previous step gives $\mathbb P(A_m)=\mathbb P(U=s_m)=0$ for every $m\in I$, hence
\begin{align*}
\mathbb P(U\in S)=0.
\end{align*}
This contradicts $\mathbb P(U\in S)=1$. Therefore no such countable full-probability set $S$ exists, so $U$ is not a discrete random variable.
[guided]
We now use the defining feature of discreteness. To say that $U$ is discrete means that all of its probability mass is carried by some countable subset of $\mathbb R$. Thus, if $U$ were discrete, there would be a countable set $S\subset\mathbb R$ satisfying
\begin{align*}
\mathbb P(U\in S)=1.
\end{align*}
First handle the possible empty support. If $S=\varnothing$, then $\{U\in S\}=\varnothing$ and therefore $\mathbb P(U\in S)=0$, which contradicts the displayed equality. Thus $S$ is nonempty.
Because $S$ is countable and nonempty, we may list its elements without repetition. More precisely, choose an index set $I$, either finite of the form $\{1,\dots,N\}$ for some $N\in\mathbb N$ or infinite of the form $\mathbb N$, together with an injective enumeration $(s_m)_{m\in I}$ such that
\begin{align*}
S=\{s_m:m\in I\}.
\end{align*}
For each $m\in I$, define the event
\begin{align*}
A_m:=\{\omega\in\Omega:U(\omega)=s_m\}.
\end{align*}
Each $A_m$ lies in $\mathcal F$ because $U$ is measurable and $\{s_m\}$ is a Borel subset of $\mathbb R$. The events are pairwise disjoint: if $m\ne n$, then $s_m\ne s_n$ by injectivity, so $U(\omega)$ cannot equal both $s_m$ and $s_n$. Moreover,
\begin{align*}
\{U\in S\}=\bigcup_{m\in I}A_m.
\end{align*}
Now apply additivity of the probability measure $\mathbb P$ to this disjoint union. If $I$ is finite, this is finite additivity; if $I=\mathbb N$, this is countable additivity. In either case,
\begin{align*}
\mathbb P(U\in S)=\sum_{m\in I}\mathbb P(A_m).
\end{align*}
The previous step proved that every singleton has probability zero under the continuous uniform distribution on $(0,1)$. Applying that result to each $s_m\in\mathbb R$ gives
\begin{align*}
\mathbb P(A_m)=\mathbb P(U=s_m)=0
\end{align*}
for every $m\in I$. Therefore
\begin{align*}
\mathbb P(U\in S)=\sum_{m\in I}0=0.
\end{align*}
This contradicts the discrete-support requirement $\mathbb P(U\in S)=1$. The contradiction shows that $U$ cannot be a discrete random variable.
[/guided]
[/step]
