[proofplan]
The proof uses conservation of the stationary probability current. Because the potential is real-valued, the current associated to any $C^2$ solution of the stationary Schrödinger equation has derivative zero, hence is constant on $\mathbb{R}$. We compute the limiting current at $-\infty$ from the incident-plus-reflected wave and at $+\infty$ from the transmitted wave. Equating these two limits gives $(\hbar k/m)(1-|R|^2)=(\hbar k/m)|T|^2$, and since $\hbar k/m>0$, the desired identity follows.
[/proofplan]
[step:Define the stationary current and prove it is constant]
Define the probability current
\begin{align*}
j:\mathbb{R}\to\mathbb{R},\qquad j(x)=\frac{\hbar}{m}\operatorname{Im}\left(\overline{\psi(x)}\,\psi'(x)\right).
\end{align*}
Since $\psi\in C^2(\mathbb{R};\mathbb{C})$, the function $j$ is $C^1$. Differentiating gives
\begin{align*}
j'(x)=\frac{\hbar}{m}\operatorname{Im}\left(\overline{\psi'(x)}\,\psi'(x)+\overline{\psi(x)}\,\psi''(x)\right).
\end{align*}
The first term $\overline{\psi'(x)}\,\psi'(x)=|\psi'(x)|^2$ is real. From the Schrödinger equation,
\begin{align*}
\psi''(x)=\frac{2m}{\hbar^2}\left(V(x)-E\right)\psi(x).
\end{align*}
Since $V(x)-E\in\mathbb{R}$, the second term satisfies
\begin{align*}
\overline{\psi(x)}\,\psi''(x)=\frac{2m}{\hbar^2}\left(V(x)-E\right)|\psi(x)|^2\in\mathbb{R}.
\end{align*}
Therefore $j'(x)=0$ for every $x\in\mathbb{R}$, so $j$ is constant on $\mathbb{R}$.
[guided]
The conserved quantity is the stationary probability current. Define
\begin{align*}
j:\mathbb{R}\to\mathbb{R},\qquad j(x)=\frac{\hbar}{m}\operatorname{Im}\left(\overline{\psi(x)}\,\psi'(x)\right).
\end{align*}
This is well-defined because $\psi:\mathbb{R}\to\mathbb{C}$ is $C^2$, hence both $\psi$ and $\psi'$ are continuous complex-valued functions. The reason this current is conserved is exactly the reality of the potential. Differentiating the product $\overline{\psi}\,\psi'$ gives
\begin{align*}
j'(x)=\frac{\hbar}{m}\operatorname{Im}\left(\overline{\psi'(x)}\,\psi'(x)+\overline{\psi(x)}\,\psi''(x)\right).
\end{align*}
The term $\overline{\psi'(x)}\,\psi'(x)=|\psi'(x)|^2$ is real, so it contributes no imaginary part. To handle the second term, solve the stationary Schrödinger equation for $\psi''(x)$:
\begin{align*}
\psi''(x)=\frac{2m}{\hbar^2}\left(V(x)-E\right)\psi(x).
\end{align*}
Because $V:\mathbb{R}\to\mathbb{R}$ is real-valued and $E\in\mathbb{R}$, the coefficient $\frac{2m}{\hbar^2}(V(x)-E)$ is real. Thus
\begin{align*}
\overline{\psi(x)}\,\psi''(x)=\frac{2m}{\hbar^2}\left(V(x)-E\right)|\psi(x)|^2
\end{align*}
is also real. Both summands inside the imaginary part are real, so $j'(x)=0$ for every $x\in\mathbb{R}$. Hence $j$ is constant. This is the analytic expression of flux conservation for a real potential: a real potential cannot create or destroy stationary probability current.
[/guided]
[/step]
[step:Compute the limiting current at $-\infty$]
Define the left asymptotic wave
\begin{align*}
\phi_-:\mathbb{R}\to\mathbb{C},\qquad \phi_-(x)=e^{ikx}+Re^{-ikx}.
\end{align*}
Its derivative is
\begin{align*}
\phi_-'(x)=ike^{ikx}-ikRe^{-ikx}.
\end{align*}
The hypotheses at $-\infty$ imply
\begin{align*}
\lim_{x\to-\infty}\left(\overline{\psi(x)}\psi'(x)-\overline{\phi_-(x)}\phi_-'(x)\right)=0.
\end{align*}
Now
\begin{align*}
\overline{\phi_-(x)}\phi_-'(x)=ik(1-|R|^2)+ik\overline{R}e^{2ikx}-ikRe^{-2ikx}.
\end{align*}
Set
\begin{align*}
z(x)=\overline{R}e^{2ikx}.
\end{align*}
Then $Re^{-2ikx}=\overline{z(x)}$, so
\begin{align*}
ik\overline{R}e^{2ikx}-ikRe^{-2ikx}=ik\left(z(x)-\overline{z(x)}\right).
\end{align*}
Since $z-\overline{z}$ is purely imaginary, $i(z-\overline{z})$ is real. Therefore the oscillatory cross term has zero imaginary part. Hence
\begin{align*}
\lim_{x\to-\infty}j(x)=\frac{\hbar k}{m}(1-|R|^2).
\end{align*}
[/step]
[step:Compute the limiting current at $+\infty$]
Define the right asymptotic wave
\begin{align*}
\phi_+:\mathbb{R}\to\mathbb{C},\qquad \phi_+(x)=Te^{ikx}.
\end{align*}
Its derivative is
\begin{align*}
\phi_+'(x)=ikTe^{ikx}.
\end{align*}
The hypotheses at $+\infty$ imply
\begin{align*}
\lim_{x\to+\infty}\left(\overline{\psi(x)}\psi'(x)-\overline{\phi_+(x)}\phi_+'(x)\right)=0.
\end{align*}
A direct computation gives
\begin{align*}
\overline{\phi_+(x)}\phi_+'(x)=\overline{T}e^{-ikx}ikTe^{ikx}=ik|T|^2.
\end{align*}
Therefore
\begin{align*}
\lim_{x\to+\infty}j(x)=\frac{\hbar k}{m}|T|^2.
\end{align*}
[/step]
[step:Equate the two flux limits and isolate the reflection-transmission identity]
Since $j$ is constant on $\mathbb{R}$, its limits at $-\infty$ and $+\infty$ are equal. Thus
\begin{align*}
\frac{\hbar k}{m}(1-|R|^2)=\frac{\hbar k}{m}|T|^2.
\end{align*}
Because $E>0$, $m>0$, and $\hbar>0$, we have $k=\sqrt{2mE}/\hbar>0$, so $\hbar k/m>0$. Dividing by $\hbar k/m$ gives
\begin{align*}
1-|R|^2=|T|^2.
\end{align*}
Rearranging yields
\begin{align*}
|R|^2+|T|^2=1.
\end{align*}
This is the claimed reflection-transmission probability conservation identity.
[/step]
