Let $l_0<l_1<l_2<l_3\in \mathbb{R}$, $I_1=(l_0,l_1),\, I_2=(l_1,l_2)$ and $I_3=(l_2,l_3)$. Define the spaces
\begin{equation*} H^2_{l_0}=\{u\in H^2(I_1):u(l_0)=u'(l_0)=0\}\\ H^2_{l_3}=\{w\in H^2(I_3):w(l_3)=w'(l_3)=0\} \end{equation*} whit the following inner products \begin{equation*} \left<u,\hat{u}\right>_{H^2_{l_0}}=\left<u'',\hat{u}''\right>_{L^2(I_1)}\\ \left<w,\hat{w}\right>_{H^2_{l_3}}=\left<w'',\hat{w}''\right>_{L^2(I_3)} \end{equation*} then, the space
\begin{equation*} \mathbb{H}:=\{(u,v,w)\in H^2_{l_0}\times H^1(I_2)\times H^2_{l_3}:u(l_1)=v(l_1) \text{ and } v(l_2)=w(l_2)\} \end{equation*}
is closed due to the continuity of the trace operator.
I have already proven that both $H^2_{l_0}$ and $H^2_{l_3}$ are closed. But I don't know how the trace operator is used to prove the rest.
Can somebody give me a hand with this?