$\newcommand{\R}{\mathbb R}\newcommand{\C}{\mathbb C} $To attach a meaning to the intersection $C:=B\cap L^2\big((0,T)\times(0,1)\big)$ of the subset $B$ of the space $L^{\infty}\big(0,T;L^1(0,1)\big)$ with the space $L^2\big((0,T)\times(0,1)\big)$, we need to identify functions $(0,T)\times(0,1)\ni(t,s)\mapsto x(t,s)\in\C$ in $L^2\big((0,T)\times(0,1)\big)$ with functions $(0,T)\ni t\mapsto x_t\in L^{\infty}\big(0,T;L^1(0,1)\big)$, which we will do by the standard formula $x_t(s):=x(t,s)$.
Let $(x^{(n)})$ be a sequence in $C$ converging to some $x\in L^2\big((0,T)\times(0,1)\big)$ in $L^2\big((0,T)\times(0,1)\big)$: \begin{equation} d_n:=\int_0^T dt\,\int_0^1 ds\, |x^{(n)}(t,s)-x(t,s)|^2\to0 \end{equation} (as $n\to\infty$). We have to show that then $x\in B$, so that $x$ is actually in $C$.
Using the Cauchy--Schwarz inequality, we have \begin{equation} \begin{aligned} \int_0^T dt\,\int_0^1 ds\, |x^{(n)}(t,s)-x(t,s)| \le\sqrt{T\times1}\Big(\int_0^T dt\,\int_0^1 ds\, |x^{(n)}(t,s)-x(t,s)|^2\Big)^{1/2} =\sqrt T\sqrt{d_n}\to0. \end{aligned} \end{equation}\begin{equation} \begin{aligned} &\int_0^T dt\,\int_0^1 ds\, |x^{(n)}(t,s)-x(t,s)| \\ &\le\sqrt{T\times1}\Big(\int_0^T dt\,\int_0^1 ds\, |x^{(n)}(t,s)-x(t,s)|^2\Big)^{1/2} =\sqrt T\sqrt{d_n}\to0. \end{aligned} \end{equation} That is, $\int_0^T dt\, g^{(n)}(t)\to0$, where $g^{(n)}(t):=\|x^{(n)}_t-x_t\|_{L^1(0,1)}$. That is, the sequence $(g^{(n)})$ of nonnegative functions converges to $0$ in $L^1(0,T)$ and hence in measure over the interval $(0,T)$. So, passing to a subsequence, without loss of generality we may assume that the sequence $(g^{(n)})$ converges to $0$ almost everywhere on $(0,T)$. That is, $\|x^{(n)}_t-x_t\|_{L^1(0,1)}\to0$ for a.a. $t\in(0,T)$. Recall that for each $n$ we have $x^{(n)}\in C\subseteq B$, whence $\|x^{(n)}_t\|_{L^1(0,1)}\le1$ for a.a. $t\in(0,T)$. So, \begin{equation} \|x_t\|_{L^1(0,1)}=\lim_n\|x^{(n)}_t\|_{L^1(0,1)}\le1 \end{equation} for a.a. $t\in(0,T)$. Thus, $x\in B$. Thus, $x\in B$. $\quad\Box$