Let $a\in L$, where $L$ is a graded (if needed) distributive lattice. Let $x_1, \ldots, x_k$ - the set of elements which cover $a$ ($x$ covers $a$ if $a < x$ and there is no element $t$ such that $a < t < x$). Denote least upper bound of $x_1, \ldots, x_k$ as $b$. Consider sublattice $[a, b]$. Is it true (and if it is, how to prove it?) that greatest lower bound of elements covered by $b$ is $a$ (in sublattice $[a,b]$)?
Let $\rho : L \to \mathbb N$ be the rank function of $L$.
Then the interval $[a,b]$ has finite length, given by $\rho(b) - \rho(a)$.
Since $[a,b]$ is a sublattice of $L$, it is distributive.
A distributive lattice is finite iff it has finite length.
Notice that the elements which cover $a$, in $L$, are the atoms of $[a,b]$;
analogously the elements covered by $b$ which are above $a$ are co-atoms of $[a,b]$.
So we're left with the task of proving that in a finite distributive lattice, if the join of the atoms is $1$, then the meet of the co-atoms is $0$.
That just follows from the fact that if the join of the atoms is $1$, then those are the only join-irreducible elements of the lattice, which is then Boolean, and it is clear that in a Boolean lattice the meet of the co-atoms is $0$.
As I suspected (and commented) the property of being graded is redundant here, and this alternative answer doesn't use it.
Notation. We write $x \prec y$ to denote that $x$ is covered by $y$, and $x \preceq y$ means that $x \prec y$ or $x = y$.
A lattice $L$ is said to satisfy the Upper Covering Condition if
$$x \preceq y \;\Longrightarrow\; x \vee z \preceq y \vee z$$
holds for every $x,y,z \in L$.
A modular lattice always satisfies the Upper Covering Condition (see G. Grätzer, General Lattice Theory, IV§1. Modular Lattices, Theorem 4).
Since a distributive lattice is modular, it satisfies the Upper Covering Condition too.
Now, let $a_1, \ldots, a_k$ be the atoms of a distributive lattice.
Form $$0 \prec a_i$$ it follows that $$x = 0 \vee x \preceq a_i \vee x,$$ for every $x \in L$. In particular, $$a_1 \preceq a_1 \vee a_2 \preceq \cdots \preceq a_1 \vee \cdots \vee a_k.$$ If additionally, we have that $a_1 \vee \cdots \vee a_k = 1$, then the lattice has finite length.
Now we follow the reasoning in the previous answer: if the lattice is distributive and has finite length then it is finite, and if it is a finite distributive lattice with the join of the atoms equal to $1$, then it is Boolean, whence the meet of the co-atoms is $0$.
Update. Actually we can circumvent the use of the Upper Covering Condition, and prove directly that, in a distributive lattice $L$, if $a$ is an atom and $x \in L$ is any other element, then $x \preceq a \vee x$ as plug in the remaining of the answer above.
Indeed, if $a \leq x$ then there is nothing to prove.
If $a \nleq x$ and there exists $c \in L$ such that $x < c < a \vee x$, then $a \nleq c$, for otherwise $a \vee x \leq c$, a contradiction; thus, as $a$ is an atom, $a \wedge c =0$, and the elements $$0,\,a,\,x,\,c,\,a\vee x$$ for a pentagon, whence the lattice is not distributive, a contradiction.