[proofplan]
We first verify that the proposed formula defines a square-integrable control on the compact interval $[0,T]$. We then compute the terminal state by the variation-of-constants representation for the linear system starting from $0$. Substituting the proposed control converts the terminal-state integral into the Gramian $W_T$ after the change of variables $s=T-t$, and nonsingularity gives $W_TW_T^{-1}x_1=x_1$.
[/proofplan]
[step:Verify that the proposed control is square-integrable]
Fix $x_1 \in \mathbb{R}^n$. Define
\begin{align*}
u: [0,T] \to \mathbb{R}^m, \qquad u(t) := B^\top e^{A^\top (T-t)} W_T^{-1} x_1.
\end{align*}
The map $t \mapsto e^{A^\top(T-t)}$ is continuous from $[0,T]$ to $\mathbb{R}^{n \times n}$, because the matrix exponential is continuous. Since multiplication by the fixed matrices $B^\top$ and $W_T^{-1}$ and by the fixed vector $x_1$ preserves continuity, $u$ is continuous on $[0,T]$. Therefore $u$ is bounded on $[0,T]$, so
\begin{align*}
\int_0^{\!T}|u(t)|^2\, d\mathcal{L}^1(t) < \infty.
\end{align*}
Thus $u \in L^2([0,T];\mathbb{R}^m)$.
[/step]
[step:Represent the terminal state using the input response formula]
Let
\begin{align*}
x: [0,T] \to \mathbb{R}^n, \qquad x(t) := \int_0^t e^{A(t-r)} B u(r)\, d\mathcal{L}^1(r).
\end{align*}
The integrand $r \mapsto e^{A(t-r)}Bu(r)$ is continuous for each fixed $t \in [0,T]$, since $u$ is continuous. Hence $x$ is the classical solution of
\begin{align*}
\dot{x}(t)=Ax(t)+Bu(t), \qquad x(0)=0.
\end{align*}
Evaluating at $t=T$ gives
\begin{align*}
x(T)=\int_0^{\!T}e^{A(T-t)}Bu(t)\, d\mathcal{L}^1(t).
\end{align*}
[guided]
We need the terminal state produced by the input $u$. For a linear system starting at the origin, the state at time $t$ is obtained by accumulating the effect of the input at each earlier time $r$, propagated forward from $r$ to $t$ by the matrix exponential. Thus define
\begin{align*}
x: [0,T] \to \mathbb{R}^n, \qquad x(t) := \int_0^t e^{A(t-r)} B u(r)\, d\mathcal{L}^1(r).
\end{align*}
This formula is legitimate because $u: [0,T]\to \mathbb{R}^m$ is continuous, the map $(t,r)\mapsto e^{A(t-r)}$ is continuous, and multiplication by the fixed matrix $B$ is continuous. Therefore the integrand is continuous on the compact region $0 \le r \le t \le T$.
The initial condition is immediate from the empty interval integral:
\begin{align*}
x(0)=0.
\end{align*}
Differentiating the integral with respect to $t$ gives the boundary contribution $Bu(t)$ from the upper limit and the derivative of the propagator inside the integral:
\begin{align*}
\dot{x}(t)=Bu(t)+\int_0^t A e^{A(t-r)}B u(r)\, d\mathcal{L}^1(r).
\end{align*}
Since $A$ is constant, it factors out of the integral:
\begin{align*}
\dot{x}(t)=Bu(t)+A\int_0^t e^{A(t-r)}B u(r)\, d\mathcal{L}^1(r).
\end{align*}
By the definition of $x(t)$, this becomes
\begin{align*}
\dot{x}(t)=Bu(t)+Ax(t)=Ax(t)+Bu(t).
\end{align*}
Thus this $x$ is exactly the trajectory of the controlled system with $x(0)=0$. Evaluating the same formula at the terminal time $T$ yields
\begin{align*}
x(T)=\int_0^{\!T}e^{A(T-t)}Bu(t)\, d\mathcal{L}^1(t).
\end{align*}
[/guided]
[/step]
[step:Substitute the Gramian control into the terminal state]
Using the definition of $u(t)$ in the terminal-state formula, we obtain
\begin{align*}
x(T)=\int_0^{\!T}e^{A(T-t)}B B^\top e^{A^\top(T-t)}W_T^{-1}x_1\, d\mathcal{L}^1(t).
\end{align*}
The vector $W_T^{-1}x_1$ is independent of $t$, so it may be factored to the right of the matrix-valued integral:
\begin{align*}
x(T)=\left(\int_0^{\!T}e^{A(T-t)}B B^\top e^{A^\top(T-t)}\, d\mathcal{L}^1(t)\right)W_T^{-1}x_1.
\end{align*}
[/step]
[step:Identify the integral with the controllability Gramian]
Apply the change of variables $s=T-t$. Under this substitution, the interval $t \in [0,T]$ is mapped to $s \in [T,0]$, and reversing the orientation gives integration over $s \in [0,T]$ with respect to $\mathcal{L}^1(s)$. Hence
\begin{align*}
\int_0^{\!T}e^{A(T-t)}B B^\top e^{A^\top(T-t)}\, d\mathcal{L}^1(t)=\int_0^{\!T}e^{As}B B^\top e^{A^\top s}\, d\mathcal{L}^1(s).
\end{align*}
By the definition of $W_T$, the right-hand side is $W_T$. Therefore
\begin{align*}
x(T)=W_TW_T^{-1}x_1=x_1.
\end{align*}
Thus the proposed control steers the system from $x(0)=0$ to the prescribed target $x(T)=x_1$.
[/step]
