Here is the question I am trying to solve:
For any Lie algebra $L$ determine the group of grouplike elements of the Hopf algebra $U(L).$
Some definitions:
1-To any Lie Algebra $L$ we assign an (associative) algebra $U(L),$ called the enveloping algebra of $L,$ and a morphism of Lie algebras $i_L: L \to L(U(L)).$ We define the enveloping algebra as follows. Let $I(L)$ be the two-sided ideal of the tensor algebra $T(L)$ generated by all elements of the form $xy - yx - [x,y]$ where $x,y$ are elements of $L.$ We define $$U(L) = T(L)/ I(L)$$
Where $U(L)$ is filtered as a quotient algebra of $T(L).$
2- Let $V$ be a vector space. Define $T^0 (V) = k, T^1(V) = V$ and $T^n(V) = V ^{\otimes n}$ (the tensor product of $n$ copies of $V$) if $n > 1.$ The canonical isomorphisms $$T^n (V) \otimes T^m(V) \cong T^{n + m}(V)$$ induce an associative product on the vector space $T(V) = \otimes_{n \geq 0} T^n (V).$ Equipped with this algebra structure, $T(V)$ is called the tensor algebra of $V.$ The product in $T(V)$ is explicitly defined by $$(x_1 \otimes \dots \otimes x_n)(x_{n + 1} \otimes \dots \otimes x_{n+m}) = x_1 \otimes \dots \otimes x_n \otimes x_{n + 1} \otimes \dots \otimes x_{n+m} \quad \quad \quad (5.1)$$
where $x_1, \dots , x_{n+1}, \dots , x_{n+m}$ are elements of $V.$ The unit for this product is the image of the unit element $1$ in $k = T^0(V).$ Let $i_V$ be the canonical embedding of $V = T^1 (V)$ into $T(V).$ By $(5.1)$ we have $$x_1 \otimes \dots \otimes x_n = i_V(x_1) \dots i_V(x_n)$$ which allows us to set $x_1 \dots x_n = x_1 \otimes \dots \otimes x_n$ whenever $x_1, \dots , x_n$ are elements of $V.$
3- A grouplike element of a coalgebra $(H, \Delta, \epsilon),$ i.e., an element $x \neq 0$ such that $$\Delta (x) = x \otimes x.$$
4- I also know that the set of grouplike elements of $H$ is a group.
Still I do not know how to answer it. Could someone show me some details please?
EDIT:
5- I also know the universal property of $U(L).$
EDIT-2:
Here is a theorem I have also but I do not know if it may help in my case or not:
Let us recall the PBW theorem. Let $\pi : T(L) \rightarrow U(L)=T(L)/I(L) $ be the canonical projection. Write $T_m = T^0(L) \oplus T^1(L) \oplus \cdots T^m(L)$ and $U_m = \pi(T_m)$. Set $U_{-1} = 0$ and $G^m = U_m/U_{m-1}$. The PBW theorem states that the graded unital associative algebra $G = \bigoplus_{m=0}^{\infty} G^m$ is isomorphic (as graded algebras) to the symmetric algebra $S(L) = \bigoplus_{m=0}^{\infty}S^m(L)$.
Let $x \in U(L)$ be a nonzero element and $x_0$ be the homogeneous component of $x$ with highest degree. In other words, $x = x_0$ mod $U_{m-1}$ where $x_0 \neq 0$ and $\deg(x_0) = m$. We will show $x^2$ has a nonzero homogeneous part of degree $2m$. According to the PBW theorem, we can consider the isomorphism of graded algebras $\psi: G= \bigoplus_{i=0}^{\infty} G^i \rightarrow S(L) = \bigoplus_{i=0}^{\infty} S^i(L)$. Here $0 \neq \overline{x} \in G^m=U_m/U_{m-1}$ is mapped to $0 \neq \psi(\overline{x}) \in S^m(L)$. Since the symmetric algebra $S(L)$ is an integral domain, $\psi(\overline{x})^2 = \psi(\overline{x^2}) \in S^{2m}(L)$ is also nonzero. Thus $\overline{x^2} \in G^{2m}$ cannot be zero.
Denote the multiplication map of $U(L)$ by $m: U(L) \otimes U(L) \rightarrow U(L)$. For every $x \in U_m$, we know $m\Delta(x)$ lies in $U_m$. This can be proved directly, but you may use the proposition in your question. However, if $x \in U_m \setminus U_{m-1}$ is a grouplike element, then $m\Delta(x)= m(x \otimes x) = x^2$ has a nonzero homogeneous part of degree $2m$. If follows that $m \geq 2m$, or $x$ is a scalar. Now $\Delta(x) = x \Delta(1)=x (1 \otimes 1)$ should be equal to $x\otimes x = x^2 (1 \otimes 1)$, whence $x = 1$.
(Added)
If $x \neq 0 $ satisfies $\Delta(x) = x \otimes x$, then $x = \mbox{id}(x) = (\epsilon \otimes \mbox{id})(\Delta x) = \epsilon(x)x$ and thus $\epsilon(x) = 1$. By the axiom of antipode $1 = xS(x) = S(x)x$, whence $x$ is invertible. From here, Mariano's answer proves the statement.