Let me describe a construction that meets your conditions.
For this, let $(P,\succeq)$ be a poset (and
always assume that $a\preceq b \Leftrightarrow b\succeq a$).
<p>


**Stage 1.** Let $(C,\succeq)$ be the
<a href="https://en.wikipedia.org/wiki/Dedekind-MacNeille_completion">Dedekind-MacNeille completion</a>
of $(P,\succeq)$, and let $h\colon P\to C\colon p\mapsto (p]$
be the order-embedding that takes an element $p\in P$
to its lower cone $(p] = \{x\in P\;|\;x\preceq p\}$.
<p>

**Stage 2.** Let $(D,\succeq)$ be the meet-subsemilattice
$(C,\succeq)$ that is generated by the image $h(P)$. If this meet-semilattice
has a least element, then label it $0$. If it does not have
a least element, adjoin one and label it $0$.<p>

At this point we have embedded $(P,\succeq)$
into a small meet-semilattice $(D,\succeq)$
with a zero element such that
the maximal elements of $(P,\succeq)$ are mapped
to the maximal elements of $(D,\succeq)$.<p>

**Stage 3.**
To each element of $d\in D$ that is not
$0$ or an atom of $(D,\succeq)$
we shall adjoin a companion element
$x_d$. Each new element $x_d$ will lie
strictly below the element $d$ and also below
any element $d'\in D$
for which $d'\succeq d$, $x_d$ will lie strictly above $0$, and
$x_d$ will be
incomparable with all other elements of $(D,\succeq)$.
Let $(X,\succeq)$ be the poset obtained from $(D,\succeq)$
by adding all companion elements.<p>

Stage 1 embeds $(P,\succeq)$ into a small complete
lattice. Stage 2 ensures that Property 2 of the
problem statement is satisfied.
Stage 2 also ensures that maximal elements are preserved under the embedding.
Stage 3 ensures that Property 1
of the problem statement is satisfied.<p>

If you apply this construction to
a chain $(P,\succeq)$ of some length $n>1$,
the resulting $(X,\succeq)$ has size $2n-2$, which is much
smaller than the size of the Boolean envelope of $P$.
If you apply this construction to
an antichain $(P,\succeq)$ of some length $n>1$,
the resulting $(X,\succeq)$ has size $n+1$, which is also much
smaller than the size of the Boolean envelope of $P$.

If $(P,\succeq)$ is a finite poset of order dimension $d$,
then it can be shown that the order dimension of the poset $(X,\succeq)$
constructed above is at most $2d \;(\leq |P|)$.
(To prove this for yourself, use the fact that the Dedekind-MacNeille completion does not increase order dimension, so the order dimension cannot increase during Stages 1 or 2.)
It was proved by Hiraguchi in 1951 that $d\leq |P|/2$
holds for any finite poset.
Thus, the order dimension of $(X,\preceq)$ is at most $2\cdot (|P|/2)=|P|$.
On the other hand, the order dimension of the Boolean envelope of
$(P,\succeq)$ is exactly $|P|$. This is another sense in which
the poset $(X,\succeq)$ is smaller than or equal to the
Boolean envelope