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?
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.