$\bigcap _{H \ne \left\{e \right\}}H \ne \left\{e \right\}$ implies every element in $G$ has a finite order Given a group $G$ and a subgroup $H \le G$, if for every $H \ne \left\{e \right\}$ : $\bigcap _{H \ne \left\{e \right\}}H$ is a subgroup different from $\left\{e \right\}$, then every element in $G$ has a finite order.

Define $$\mathcal H := \bigcap _{H \ne \left\{e \right\}}H$$
Since $\mathcal H$ is the  intersection of a nonempty family of subgroups of $G$  hence $\mathcal H \le G$,from the definition of intesction it follows that there is another $h \ne e$ such that $h \in \mathcal H$,the cyclic group $\langle h \rangle$ is a subgroup of $\mathcal H$ and from $\langle h \rangle \subseteq \mathcal H \subseteq G$ and the definition of $\mathcal H$ we conclude that  $\mathcal H$ is a subset of $\langle h \rangle$ ,Implies $\mathcal H=\langle h \rangle$.
Clearly $2 \le\text{ord}(\mathcal H )$,if the order is $2$ then $h^2=e$,otherwise we can construct a cyclic subgroup $\langle h^2 \rangle \subseteq \mathcal H$ from the previous arguments $\mathcal H= \langle h^2 \rangle$,on the other hand $h \in \langle h \rangle = \langle h^2 \rangle$ so there is some integer $k$ for which $h=h^{2k}$ if and only if $h^{2k-1}=e$.
So it's shown that $\mathcal H$ is a finite cyclic subgroup of $G$.
If $g \ne e$ then the subgroup generated by $g$ is a nontrivial subgroup and by the definition of $\mathcal H$: $h \in \langle h \rangle= \mathcal H \subseteq\langle g \rangle$,So there exists an integer $j$ such that $g^j=h$.
If we denote by $m$ the order of $\mathcal H$,then $(g^j)^m=h^m=e$
So every non-identity element in $G$ has a finite order (and $g=e$ has order $1$) ..

But there is a problems,we know that $k$ was an integer and $h^{2k-1}=e$ ,this is true to claim $\text{ord}(h)=2k-1$ as long as $2k-1$ is a positive integer,How do we ensure that?
And we know that $\text{ord}(g)=jm$, ,this is true as long as $jm$ is a positive integer,But as it's clear $j$ can be a negative inetegr and hence $jm$ would be a negative integer and hence cannot be the order of an element (since the order of an element is always a positive integer).
So how can someone explain that?
 A: Your problems aren't really problems. In general, if $x^n=e$ then $x^{-n}=e^{-1}=e$.
But please note that showing $x^n=e$ doesn't necessarily mean $ord(x)=n$, even when $n$ is positive. It just means that $ord(x)$ divides $n$.
In conclusion, if you know $x^n=e$, and $n\neq 0$,  then you can conclude that $ord(x)$ is finite and divides $|n|$.
So in your proof you know that $ord(h)$ divides $|2k-1|$, and later you know that $ord(g)$ divides $|jm|$. But this is good enough for what you want.

Final remark on writing. The way you have written the problem is still confusing/incorrect. For example, in the title you write
$$
\forall \{e\}\neq H\leq G : \bigcap_{H\neq\{e\}}H\neq \{e\}
$$
If I rephrase this in words it says: "For every nontrivial subgroup of $G$, the intersection of all nontrivial subgroups of $G$ is nontrivial." So the point is that you don't need to say "$\forall \{e\}\neq H\leq G$". The assertion "$\bigcap_{H\neq \{e\}}H\neq\{e\}$" is a complete phrase on it's own (as long as it's understood from context that the letter $H$ refers to subgroups, otherwise you might write $\bigcap_{\{e\}\neq H\leq G}H$ instead).
For a comparison, this would be similar to writing "for all $n\geq 1$, $\lim_{n\to \infty}\frac{1}{n}=0$" instead of just "$\lim_{n\to\infty}\frac{1}{n}=0$". The first statement is confusing, while the second statement is complete and correctly written.
