(Building on Goldstern's comment:) If fields are ok, then the distributivity certainly does not hold.

Take e.g. a finite degree extension $F$ of $\mathbb{Q}$ with Galois group $G$. Subalgebras of $F$ are subfields, by undergraduate field theory, so the subalgebra lattice over $\mathbb{Q}$ is the field extension lattice of $F:\mathbb{Q}$. (I'm assuming that you're taking all your algebras over a fixed field, here $\mathbb{Q}$.) By the Galois correspondence, the field extension lattice is anti-isomorphic to the subgroup lattice of the Galois group.

And subgroup lattices certainly need not be distributive. Indeed, by a theorem of Ore, the subgroup lattice $L(G)$ of a finite group $G$ is distributive iff $G$ is cyclic.

(Building on Goldstern's comment:) If fields are ok, (and if you allow infinite algebras -- see comment by Mariano Suárez-Alvarez below) then the distributivity certainly does not hold.

Take e.g. a finite degree extension $F$ of $\mathbb{Q}$ with Galois group $G$. Subalgebras of $F$ are subfields, by undergraduate field theory, so the subalgebra lattice over $\mathbb{Q}$ is the field extension lattice of $F:\mathbb{Q}$. (I'm assuming that you're taking all your algebras over a fixed field, here $\mathbb{Q}$.) By the Galois correspondence, the field extension lattice is anti-isomorphic to the subgroup lattice of the Galois group.

And subgroup lattices certainly need not be distributive. Indeed, by a theorem of Ore, the subgroup lattice $L(G)$ of a finite group $G$ is distributive iff $G$ is cyclic.