Energy method for one dimensional wave equation with Robin boundary condition Show that the initial-boundary value problem   
\begin{align}
  & {{u}_{tt}}={{u}_{xx}}\text{    }(x,t)\in \left( 0,l \right)\times \left( 0,T \right),\text{ }T,l>0 \\ 
 & u\left( x,0 \right)=0,\text{   }x\in \left[ 0,l \right] \\ 
 & {{u}_{x}}\left( 0,t \right)-u\left( 0,t \right)=0,\text{ }{{u}_{x}}\left( l,t \right)+u\left( l,t \right)=0,\text{   }t\in \left[ 0,T \right]\\ 
\end{align}  
has zero solution only.  
My attempt 2: 
Previously I tried separation by variables but got stuck at the end.  Inspired by BCLC, I try energy method this time.  
Set  
$$E\left( t \right)=\frac{1}{2}\int_{0}^{L}{\left( u_{x}^{2}\left( x,t \right)+u_{t}^{2}\left( x,t \right) \right)dx}.$$
The equation ${{u}_{tt}}={{u}_{xx}}$ and the Robin b.c. gives  
$$\begin{align}
  & \frac{dE}{dt}=\int_{0}^{L}{\left( {{u}_{x}}{{u}_{xt}}+{{u}_{t}}{{u}_{tt}} \right)dx} \\ 
 & \text{     }=\int_{0}^{L}{\left( -{{u}_{t}}{{u}_{xx}}+{{u}_{t}}{{u}_{tt}} \right)dx}+\left. {{u}_{t}}{{u}_{x}} \right|_{0}^{L} \\ 
 & \text{     }={{u}_{t}}\left( l,t \right){{u}_{x}}\left( l,t \right)-{{u}_{t}}\left( 0,t \right){{u}_{x}}\left( 0,t \right) \\ 
 & \text{     }=-{{u}_{t}}\left( l,t \right)u\left( l,t \right)-{{u}_{t}}\left( 0,t \right)u\left( 0,t \right)\le 0\text{  }\left( ? \right) \\ 
\end{align}$$  
Therefore, $E\left( t \right)\le E\left( 0 \right)$ for all $t\ge 0$. Since $E\left( t \right)\ge 0$ and  , we obtain $E\left( 0 \right)=0 (?)$ for all $t\ge 0$, thus $E\equiv 0$ and hence $u\equiv 0$ .  
Is the proof correct?

$\newcommand{\e}{\epsilon}$The first $(?)$ is indeed questionable. Instead, note that you have shown that
$$E'(t) = -u_{t}(l, t)u(l, t) -u_{t}(0, t)u(0, t). \tag{$1$}$$
Defining $\e : [0, \infty) \to \Bbb R$ by $$\e(t) := \frac{1}{2}[u^{2}(l, t) + u^{2}(0, t)],$$
we see that $(1)$ tells us that $(E + \e)' \equiv 0$.
Moreover, $E(0) = \e(0) = 0$. Thus, $E + \e$ is identically equal to $0$. Since $\e$ and $E$ are both nonnegative functions, it follows that $E \equiv 0$. Now, you can conclude $u \equiv 0$ as before.

In the above, I'm assuming that we are also given the data $u_t(x, 0) = 0$ for $x \in [0, l]$. That is how I'm concluding $E(0) = 0$.
