Showing $\tilde\chi: G/N \rightarrow \mathbb{C} $ is a character of group $G/N$. So I need to prove the following:

Let $\chi:G\rightarrow \mathbb{C}$ be a character of group $G$, with the property that $N \leq \ker \chi$, show that:
  $$\tilde\chi: G/N \rightarrow \mathbb{C} \qquad \tilde\chi(gN)=\chi(g) $$
is a character of the $G/N$ group.

My attempt:
Let's choose a representation $\rho$ of $G$,  $\rho:G\rightarrow GL(V)$, where $V$ is a finite dimensional vector space over a field. Since $N\leq \ker \chi\leq G$ means that for $n \in N \implies n \in G$ and also for $gn \in G/N \implies gn \in G$ (due to closure of $G$). So we can use the same representation for $G/N$ as well. 
For $\tilde\chi$, as defined above, to be a character the following needs to hold:
$$\tilde \chi_\rho(gn)=Tr{\{\rho(gn)} \},$$
for all $gn\in G/N$.
The kernel of $\chi$ is the set: $\ker \chi_\rho=\{g \in G |\chi_\rho(g)=\chi_\rho(id)\}$, for the representation $\rho$ of $G$. For $g \in \ker \chi$, we  have that $\rho(g)=S\rho(id)S^{-1}$,where $S$ is some invertible matrix, since both $\rho(g)$ and $\rho(id)$ have the same character (trace). 
Now we can write:
 $$\chi_\rho(gn)=Tr \{\rho (gn)\}=Tr \{ \rho(g)\rho(n) \}$$
$$=Tr\{ \rho(g) S \rho(id) S^{-1} \}=$$
$$\text{using the fact that:} \space \rho(id)=1\quad \text{(from representation theory)}$$
$$=Tr\{\rho(g)\}=\chi_\rho(g).$$
We have therefore shown that $\forall gn \in G/N:\chi_\rho (gn)=\chi_\rho(g)$ and therefore if $\tilde\chi_\rho$ is given by: $\tilde\chi_\rho(gN)=\chi_\rho(g)$, then $\tilde\chi$ is indeed a character of $G/N$.
My question:
Could some please provide some input on my attempt, I would really appreciate some help. I double-checked everyting and it seems allright. Maybe someone could provide a different approach to this.
 A: I think your attempt is mostly correct if $G$ is assumed to be finite (which I suspect to be the case) and the ideas you employ make sense to me then. I just want to point out some points where I think there might be some confusion or more precision might be useful.
It is true that two representations $\rho_1$ and $\rho_2$ of a finite group $G$ are equivalent (i.e. there exists an invertible matrix $S$ such that $\rho_1(g) = S \rho_2(g) S^{-1}$ for all $g \in G$) if and only if the two characters $\chi_{\rho_1}$ and $\chi_{\rho_2}$ of $\rho_1$ and $\rho_2$ coincide (i.e. $\chi_{\rho_1}(g) = \chi_{\rho_2}(g)$ for all $g \in G$). This is a very important result in the character theory of finite groups. 
You should compare this with your own claim in your question: You said that having $Tr(\rho(g)) = Tr(\rho(id))$ implies $\rho(g) = S \rho(id) S^{-1}$ for some invertible $S$. While this might be true in this case, referring to the above statement (identical character implies equivalent representation) as the reason for this is incorrect since you are not comparing two characters at each $g \in G$ but two values of a single character.
I have emphasized in the word "finite" in my previous statements and, in fact, the way in which you want to prove the statement only works for finite groups: You want to show/use that for any $g \in \ker(\chi)$ you already have $\rho(g) = \rho(id)$ (as you mention in the comments. Also note there is no need for any $S$ since $\rho(id)$ is the identity matrix). 
This, however, is only a fact for finite groups but not for infinite ones, for example the representation $$\rho : \mathbb{Z} \to GL_2(\mathbb{C}), n \mapsto \left(\begin{matrix} 1 & n \\ 0 & 1 \end{matrix}\right)$$ satisfies $Tr(\rho(n)) = Tr(\rho(id))$ for all $n \in \mathbb{Z}$ but $\rho(n) \neq \rho(0)$ for $n \neq 0$.  
For finite groups, the reason why this works comes from the fact that $\rho(g)$ has finite order, so $\rho(g)$ is diagonalisable and its eigenvalues are all roots of unity. The trace of $\rho(g)$ is the sum of its eigenvalues and in order for $n$ roots of unity to add up to $n$ it is necessary and sufficient that these roots of unity are all $1$.
Given that fact, you have proved that $\widetilde{\chi}$ is actually well-defined (i.e. $\chi$ is constant on $N$-cosets).
But if one wants to be rigorous, then you should also prove that $\widetilde{\chi}$ is a character of $G/N$, that is, show that $\widetilde{\chi} = \chi_{\widetilde{\rho}}$ is the character for a representation $\widetilde{\rho}$ of $G/N$. This can be established by noting that $\rho: G \to GL(V)$ with $N \subseteq \ker(\rho)$ induces a homomorphism $\widetilde{\rho} : G/N \to GL(V)$ 
with $\rho(g) = \widetilde{\rho}(gN)$ for all $g \in G$. When you said "so we can use the same representation for $G/N$", this might be what you had in mind but for this to work you have to prove $\ker(\rho) = \ker(\chi_\rho)$ first.
At the beginning of your attempt, your statement "$gn \in G/N \implies gn \in G$" appears strange to me. Keep in mind that the group $G/N$ is not a subset of $G$, its elements are.
