Implications $1.\Rightarrow 2.\Rightarrow 3.$ are clear. We will show $3.\Rightarrow 2.\Rightarrow 1.$

$(3.\Rightarrow 2.)$ If $A=\varnothing$, then $\bigvee A=0$ by definition. So assume $A$ be a non-empty subset of $L$. Let $A^{{\prime}}$ be the set of all finite subsets of $A$ and $B=\{\bigvee F\mid F\in A^{{\prime}}\}$. By assumption, $B$ is well-defined and $A\subseteq B$. Next, let $B^{{\prime}}$ be the set of all directed subsets of $B$, and $C=\{\bigvee D\mid D\in B^{{\prime}}\}$. By assumption again, $C$ is well-defined and $B\subseteq C$. Now, every chain in $C$ has a maximal element in $C$ (since a chain is a directed set), $C$ itself has a maximal element $d$ by Zorn’s Lemma. We will show that $d$ is the least upper bound of elments of $A$. It is clear that each $a\in A$ is bounded above by $d$ ($A\subseteq B\subseteq C$). If $t$ is an upper bound of elements of $A$, then it is an upper bound of elements of $B$, and hence an upper bound of elements of $C$, which means $d\leq t$.

$(2.\Rightarrow 1.)$ By assumption $\bigvee\varnothing$ exists ($=0$), so that $\bigwedge L=0$. Now suppose $A$ is a proper subset of $L$. We want to show that $\bigwedge A$ exists. If $A=\varnothing$, then $\bigwedge A=\bigvee L=1$ by definition of an arbitrary meet over the empty set. So assume $A\neq\varnothing$. Let $A^{{\prime}}$ be the set of lower bounds of $A$: $A^{{\prime}}=\{x\in L\mid x\leq a\mbox{ for all }a\in A\}$ and let $b=\bigvee A^{{\prime}}$, the least upper bound of $A^{{\prime}}$. $b$ exists by assumption. Since $A$ is a set of upper bounds of $A^{{\prime}}$, $b\leq a$ for all $a\in A$. This means that $b$ is a lower bound of elements of $A$, or $b\in A^{{\prime}}$. If $x$ is any lower bound of elements of $A$, then $x\leq b$, since $x$ is bounded above by $b$ ($b=\bigvee A^{{\prime}}$). This shows that $\bigwedge A$ exists and is equal to $b$. ∎