if we have ({1},>=)
then 1>=1 therefore reflexive ,anti-symmetric as (1,1) ,transitive . so it is a POSET and 1^1=1 and again 1LUB1=1 so 1 is the complement of itself the set has LUB and GLB for every pair of element and hence a lattice and it is a boolean algebra therefore the smallest lattice contains only one element .
One of my friend suggest me this approach. is this approach correct ? if not please suggest some method to prove that it's wrong .
This is right. More generally, any set of purely equational axioms is satisfied in the one-element structure in its language.$^1$ For example, this applies to groups, rings, Boolean algebras, lattices, ... The one-element [whatever] is generally called the trivial [whatever], e.g. trivial group, trivial ring, trivial Boolean algebra, trivial lattice, ...
By "purely equational" above I mean that the axioms should all have the form $$\forall x_1,...,x_n(t_1(x_1,...,x_n)=t_2(x_1,...,x_n))$$ where $t_1,t_2$ are terms in the relevant language. Note that disjunctions, inequalities, and existential quantifiers aren't allowed in this form. For example, the axiom of associativity in groups has this form with $n=3$, $t_1(x_1,x_2,x_3)=(x_1*x_2)*x_3$, and $t_2(x_1,x_2,x_3)=x_1*(x_2*x_3)$: $$\mbox{Associativity } \equiv\forall x_1,x_2,x_3((x_1*x_2)*x_3=x_1*(x_2*x_3)).$$
It's important to note that not all kinds of structure are given by purely equational axioms! The standard example is the class of fields. There are two key field axioms which are non-equational:
$0\not=1$.
$\forall x(x=0$ or $x\cdot x^{-1}=1)$.
$^1$Actually I'm lying here a bit. There are two main ways to describe what a Boolean algebra is: either it's a poset satisfying certain (non-equational) axioms, or it's a set equipped with two binary operations $\wedge$, $\vee$, a unary operation $'$, and two constants $\top$, $\perp$ satisfying certain (equational) axioms. You've referred to the former presentation, which is not equational, so strictly speaking this result doesn't apply.
However, these two presentations of Boolean algebras are in fact equivalent in an appropriate sense: from one presentation we can recover the other in a "definable" way (this can be made precise via model theory, but I'm not going to go into that here). Specifically:
To go from the poset to the algebra, we just interpret $\wedge$ as greatest lower bound, $\vee$ as least upper bound, $\top$ and $\perp$ as top and bottom element, and $'$ as complement in the sense of $\le$.
To go from the algebra to the poset, we just interpret $\le$ by setting $x\le y\iff x\wedge y=x$.
In each case, the axioms on one "side" ensure that the interpretation from that "side" to the other "side" makes sense and and behaves as desired. (This is a good exercise!)