If $L$ is an algebraic lattice with a complete sublattice $T$, must $T$ be algebraic? I suspect that it must not be, but am struggling to find a natural counterexample.
As supremums are preserved in $T$, compact elements of $L$ in $T$ are compact in $T$. Thus, if $T$ is to not be compactly generated, $T$ must not contain some compact elements of $L$. However, in each of my cursory attempts at formulating $T$, new elements became compact in $T$ relative to $T$, so that $T$ was once again compactly generated.
Let $L$ be a complete lattice and let $T$ is a subset of $L$ that is closed under infima. Then $T$ will also have suprema, though the inclusion $T \hookrightarrow L$ need not preserve them. Define $f : L \to T$ by $$f (x) = \inf {\{ t \in T : x \le t \}}$$ By construction, $x \le f (x)$, and $f (x) \le t$ if and only if $x \le t$. Hence, $f : L \to T$ preserves suprema.
Now suppose the inclusion $T \hookrightarrow L$ preserves directed suprema. Then $f : L \to T$ also preserves compact elements. So if $x \in L$ is the supremum (in $L$) of the compact elements (in $L$) under it, then $f (x)$ is also the supremum (in $T$) of the compact elements (in $T$) under it. Thus, we have proved:
Proposition. If $L$ is an algebraic lattice and $T$ is a subset of $L$ that is closed under infima and directed suprema, then $T$ is also an algebraic lattice.