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Yes, that is true. This is a consequence of the classical Trace Theorem for Sobolev spaces. The proof can be found in any textbook on Sobolev spaces for example in Evans' Partial Differential Equations. The usual statement of the trace theorem deals with traces of $T:W^{1,p}(\Omega)\to L^p(\partial\Omega)$, but the proof would also give traces $T:W^{1,p}([0,T]\times Y)\to L^p(\{ t\}\times Y)$. If $u\in W^{2,p}$, then the derivative is in $W^{1,p}$ so the trace of the derivative is in $L^p$ and since the derivative of the trace is the trace of the derivative (roughly speaking) we obtain the trace $T:W^{2,p}([0,T]\times Y)\to W^{1,p}(\{ t\}\times Y)$.