If $Y$ is a subgroup of a topological group $(X,*,\cal T)$ then $Y$ is open if and only if $\operatorname{int}Y$ is not empty. 
Definition
A topological group is a group $(X,*)$ equipped with a topology $\cal T$ with resepct the functions
$$
p:X\times X\ni (x_1,x_2)\longrightarrow x_1*x_2\in X\quad\text{and}\quad s:X\ni x\longrightarrow x^{-1}\in X
$$
are continuous.

Let be $(X,*,\cal T)$ a topological group. So first of all we prove that if $y\in\operatorname{int}Y$ with $Y\mathcal P(X)$ then there exists a neighborhood $V_e$ of the identity element $e$ such that
$$
y*V_e\subseteq Y
$$
So first of all we observe that the identity
$$
e*y=y
$$
holds for any $y\in\operatorname{int}Y$ so that we conclude that
$$
(e,y)\in p^{-1}[\operatorname{int}Y]
$$
Now $p$ is continuous so that $p^{-1}[\operatorname{int}Y]$
there exists open neighborhoods $A_e$ and $A_y$ of $e$ and $y$ respectively such that
$$
(e,y)\in A_e\times A_y\subseteq p^{-1}[\operatorname{int}Y]
$$
that is such that
$$
y=e*y\in (A_e*A_y)\subseteq p\big[p^{-1}[\operatorname{int}Y]\big]\subseteq\operatorname{int}Y
$$
So through the identity
$$
e*y=y
$$
we conclude that
$$
y\in(y*A_e)\subseteq A_y*A_e\subseteq\operatorname{int}Y\subseteq Y
$$
as we desired.
Now we let to prove that a subgroup $Y$ of $X$ is open if and only if its interior $\operatorname{int}Y$ is not empty. So if $\operatorname{int}Y$ is not emptyset then there exists $y_0\in\operatorname{int}Y$ so that there exists a neighrborhood $V_e$ of $e$ such that
$$
y_0*V_e\subseteq Y
$$
Now $Y$ is a group and so if $y\in Y$ then we observe that
$$
y*Y\subseteq Y
$$
so that by the inclusion
$$
y_0*V_e\subseteq Y
$$
we conclude that
$$
y*V_e=(y*y_0^{-1})*y_0*V_e\subseteq y*Y\subseteq Y
$$
and thus if $y*V_e$ was open the statement follows but unfortunately, I believe that this is not true although my topology text suggests to use the last inclusion. So how prove the statement? pheraps is it false? could someone help me, please?
 A: To follow the suggested proof by azif00
First of all we prove that the function
$$
f_{x_0}:X\ni x\longrightarrow x_0*x\in X
$$
is a homeomorphism for any $x_0\in X$.  So first of all we observe that the map
$$
\iota_0:X\owns x\longrightarrow(x_0,x)\in X\times X
$$
is an embedding so that by the identity
$$
f_{x_0}=p\circ\iota_0
$$
we conclude that $f_{x_0}$ is continuous and thus by the identity
$$
f_{x_0}\circ f_{x_0^{-1}}=\text{id}_X=f_{x_0^{-1}}\circ f_{x_0}
$$
we finally conclude that $f_{x_0}$ is a homeomorphism. Now we suppose that for $y_0\in Y_0$ with $Y_0\in\mathcal P(X)$ there exists a neighborhood $N_e$ of $e$ such that
$$
y_0*N_e\subseteq Y_0
$$
So through the identity
$$
f_{y_0}[N_e]=y_0*N_e
$$
we observe that $y_0*N_e$ is a neighborhood of $y_0$ so that if the last inclusion holds then $y_0$ is an interior point of $Y_0$ so that by shown into the question we conclude that a point $y_0\in Y_0$ is an interior point if and only if there exists a neighborhood $N_e$ of $e$ such that
$$
y_0*N_e\subseteq Y_0
$$
So by what shown we conclude that the inclusion
$$
y*V_e
$$
proves that any $y\in Y$ is a interior point so that $Y$ is open.
