Let $L$ be a lattice and $x\in L$ a fixed element. Consider $L':=\{y\in L\mid x \leq y\}$.
Is $L'$ a lattice?
If $a,b\in L'$, then $x\leq a \leq a\lor b\in L'$, so I guess the join in $L'$ is just the join in $L$.
But I'm not sure whether $L'$ has a meet operation. Let again $a, b\in L'$. Then the meet of $a, b\in L$ is an element $a\land b\in L$ such that $a\land b\leq a$ and $a\land b\leq b$. I don't see whether I conclude from this that $a\land b$ are in $L'$. But $a\land b\in L'$ should be true if the implication $$(x\leq a) \,\&\, (x\leq b)\implies (x\leq a\land b)$$ holds in a lattice, because then by definition of $L'$ we have $a\land b\in L'$.
Does this implication hold in any lattice?
A further question is:
If $L$ is a complete lattice, is $L'$ a complete lattice?
$L'$ is a lattice since $x \leq a$ and $x \leq b$ implies that $x \leq a \wedge b$ (which is one of your questions).
Again, if $L$ is complete, and $A \subseteq L'$, then $x \leq \bigwedge A$, by the same reason.
So $L'$ will be complete too (having $x$ as its bottom element).