= Solution
Compact support and the <fundamental theorem of calculus> give, for every $(x,y)$,
$$
\begin{aligned}
|u(x,y)|^2
&=\left|\int_{-\infty}^y\partial_t|u(x,t)|^2\,dt\right|\\
&\leq2\int_{\mathbb R}|u(x,t)|\,|\partial_2u(x,t)|\,dt
\leq2A(x),
\end{aligned}
$$
where
$$
A(x)=\int_{\mathbb R}|u(x,t)|\,|\nabla u(x,t)|\,dt.
$$
Applying the same argument in the other coordinate gives
$$
|u(x,y)|^2\leq2B(y),
\qquad
B(y)=\int_{\mathbb R}|u(t,y)|\,|\nabla u(t,y)|\,dt.
$$
Multiplication and <Fubini's theorem> now yield
$$
\begin{aligned}
\int_{\mathbb R^2}|u(x,y)|^4\,dx\,dy
&\leq4\int_{\mathbb R^2}A(x)B(y)\,dx\,dy\\
&=4\left(\int_{\mathbb R^2}|u|\,|\nabla u|\right)^2.
\end{aligned}
$$
Finally, the <Cauchy-Schwarz inequality> gives
$$
\left(\int_{\mathbb R^2}|u|\,|\nabla u|\right)^2
\leq
\left(\int_{\mathbb R^2}|u|^2\right)
\left(\int_{\mathbb R^2}|\nabla u|^2\right).
$$
\b[Thus the required estimate holds with $C=4$; it is the <Ladyzhenskaya inequality in two dimensions>.]
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