Under the map
$$\phi:R\rightarrow R/N$$ $$r\rightarrow r+N$$ where $R$ is a field and $N$ is an ideal of $R$. it is a well known facts that the image of an ideal is again an ideal and the inverse image of an ideal is again an ideal(under a homomorphism).
now why would the image under $\phi$ of a proper non trivial ideal that properly contains $N$ is again a proper non trivial ideal.
Let $I$ be an ideal of $R$ properly containing $N$, i.e. $N \subseteq I$ and $N \neq I$. Let $J \subseteq R/N$ be the image of $I$ under the projection $\pi: R \to R/N$. To see that $J \neq 0$, choose $x \in I-N$. Then $\pi(x) \neq 0$ in $R/N$ since $x+N=0+N$ implies that $x \in N$. To see that $J \neq R/N$, I claim that $\pi(1) \notin J$. If it was we could write $1=r+n$ for some $r \in I$ and $n \in N$. But $N \subseteq I$ which implies $1 \in I$, a contradiction.