[step:Prove uniqueness in the whole energy class on the same time interval]
Let $u,v\in X_T$ be two solutions of the semilinear wave equation with the same initial data. Since $u,v\in X_T$, define
\begin{align*}
M:=\|u\|_{X_T}+\|v\|_{X_T}.
\end{align*}
The difference $z:=u-v$ belongs to $X_T$, has zero initial data at $t=0$, and satisfies
\begin{align*}
\partial_t^2z-\Delta z=-(F(u)-F(v))
\end{align*}
in $\mathcal{D}'(\mathbb{R}^n\times I_T)$. The forcing term belongs to $L^1(I_T;L^2(\mathbb{R}^n))$ by the local Lipschitz estimate for $F$ and the bounds defining $M$, so the shifted energy estimate for energy-class distributional solutions applies to $z$ on every closed subinterval of $I_T$.
Let $J=[a,b]\subset I_T$ be a closed interval on which $z(a)=0$ and $\partial_tz(a)=0$ in $E$, and define
\begin{align*}
\|z\|_{J}:=\sup_{t\in J}\bigl(\|z(t)\|_{H^1(\mathbb{R}^n)}+\|\partial_tz(t)\|_{L^2(\mathbb{R}^n)}\bigr).
\end{align*}
For every $s\in J$, the nonlinear Lipschitz estimate gives
\begin{align*}
\|F(u(s))-F(v(s))\|_{L^2(\mathbb{R}^n)}\leq L_{n,p}M^{p-1}\|z(s)\|_{H^1(\mathbb{R}^n)}.
\end{align*}
Applying the shifted forced wave estimate with initial time $a$ and zero data gives
\begin{align*}
\|z\|_{J}\leq C_{\mathrm{w}}(b-a)L_{n,p}M^{p-1}\|z\|_{J}.
\end{align*}
Choose $\tau_0>0$ such that
\begin{align*}
C_{\mathrm{w}}\tau_0L_{n,p}M^{p-1}<1.
\end{align*}
Then the preceding estimate forces $\|z\|_{J}=0$ on every interval $J=[a,b]\subset I_T$ of length at most $\tau_0$ whose left endpoint has zero Cauchy data for $z$.
Starting with $a=0$, we obtain $z=0$ on $[0,\min\{\tau_0,T\}]$. At the right endpoint of this interval, continuity of $z\in C(I_T;H^1(\mathbb{R}^n))$ and $\partial_tz\in C(I_T;L^2(\mathbb{R}^n))$ gives zero Cauchy data again, so the same estimate applies to the next interval of length at most $\tau_0$. Since finitely many intervals of length at most $\tau_0$ cover $[0,T]$, induction gives $z=0$ on $[0,T]$.
The same argument run backward in time, using intervals $[b,a]$ with $b<a$ and the shifted estimate integrated from $a$ down to $b$, gives $z=0$ on $[-T,0]$. Hence $z=0$ on all of $[-T,T]$, so the solution is unique in $X_T$.
[/step]