It is easy to show that if $\varphi: L \to M$ is a complete lattice homomorphism, then $\varphi(L)$ is a complete sublattice of $M$. It follows that, under the same assumption, if $L'$ is a complete sublattice of $L$, then $\varphi(L')$ is a complete sublattice of $M$ (and of $\varphi(L)$).
We also know that a subset $L'\subseteq L$ can be a complete lattice under the order relation of $L$ without being a complete sublattice of $L$.
Question: does the following assertion hold? If $\varphi: L \to M$ is a complete lattice homomorphism, if $L' \subseteq L$ is a complete lattice, then $\varphi(L')$ is a complete lattice.
This is false; indeed, $\varphi(L')$ may not even be a lattice. For instance, let $L$ be the free Boolean algebra on four elements $a,b,c,d$, and let $M$ be a finite Boolean algebra with four distinct elements $x,y,z,w$ such that $x,y\leq z,w$ but they are otherwise incomparable. Let $\varphi:L\to M$ be the unique Boolean homomorphism that maps $a,b,c,d$ to $x,y,z,w$. It is a complete homomorphism since $L$ and $M$ are finite.
Now let $L'=\{0,a,b,c,d,1\}\subset L$, which is a complete lattice since $a,b,c,d$ are incomparable. Then $\varphi(L')=\{0,x,y,z,w,1\}$, which is not a lattice since $x$ and $y$ have no join.