If $N \leq M$ and $Rad(M/N)=\{ 0\}$, then $Rad(M)\subseteq N$.
As $Rad(M/N)=\{0\}$, by definition, $$\bigcap\{T\leq M/N\;|\;(M/N)/T \;is\;simple\}=\{0\}$$ But how I make the link between $Rad(M)$ and $N$.
So $\{0\} \in \{T\leq M/N\;|\;(M/N)/T \;is\;simple\}$, then can I say that $\frac{M/N}{N/N} \cong M/N$ is simple.
If so, then by the correspondence theorem there is no proper submodule that contains $N$ in correspondence with the submodules of the quotient, then as $Rad(N)$ is a submodule of $M$ then $Rad(M) \subseteq N$?
Firstly, let's be careful about something in here: $\cap_{i} A_i=\{0\}$ does not imply necessarily that $A_j=\{0\}$ for some $j$, so your argument saying that $\{0\}$ must be in that set is not correct. As an example, $Rad(\mathbb{Z}_{\mathbb{Z}})=\cap_{p\in P} \mathbb{Z}/p\mathbb{Z}=\{0\}$, with $P$ being the set of all prime numbers.
Now notice that if we have two modules $A\leq B$, then $B/A$ is a simple module if and only if $A$ is a maximal submodule of $B$ (i.e. there is no proper submodule of $B$ containing $A$ properly).
Also, every maximal submodule of $B/A$ has the form $K/A$ for some maximal submodule $A\leq K\leq B$.
Therefore, going back to your case we can say: \begin{equation} 0=Rad(M/N)=\cap \{M_i: M_i~ \text{maximal submodule of}~ M/N\}=\cap \{K_i/N: N\leq K_i\leq M~ \text{maximal submodule of}~ M\} \end{equation}
But this means that $\cap_{i} K_i=N$, and so since $Rad(M)$ is the intersection of all the maximal submodules of $M$ and every $K_i$ is a maximal submodule we conclude that $Rad(M)\leq \cap_{i} K_i=N$.