Frattini subgroup is normal-monotone In the exercise 6.1.22 of Dummit and Foote's Abstract Algebra (Here $\Phi(G)$ is the Frattini subgroup of $G$):

If $N\unlhd G$, then $\Phi(N)\subseteq\Phi(G)$.

When every proper subgroup of $N$ is contained in a maximal subgroup of $N$, the statement can be proved. (By taking $M$ as a maximal subgroup of $G$ that fails to contain $\Phi(N)$ and deriving $N=\Phi(N)(N\cap M)$, a contradiction.)
But, as there may not exist a maximal subgroup of $N$ containing $N\cap M$, the case is different when some proper subgroup of $N$ is not contained in a maximal subgroup of $N$. 
Hence I'd like to ask that if $N\unlhd G$ and $M$ a maximal subgroup of $G$, could the case that there does not exist a maximal subgroup of $N$ containing $N\cap M$ happen? Or is there another way to prove the statement? (Otherwise, is there a counterexample that the statement does not hold when some proper subgroup of $N$ is not contained in a maximal subgroup of $N$?)
In fact, I wonder that if a group satisfies the condition that every proper subgroup is contained in a maximal subgroup, could it be possible that the condition does not apply to its normal subgroup?
 A: partial answer: Thanks to a comment I realise my answer is only sufficient in the case $\Phi(N)$ is finitely generated.
You don't need every proper subgroup of $N$ to be contained in a maximal subgroup of $N$ to reach $N=\Phi(N)(N\cap M)$. 
If $M$ is a maximal subgroup of $G$ not containing $\Phi(N)$ then $G=\Phi(N)M$ so by the modular law for groups we have $$N=N\cap\Phi(N)M=\Phi(N)(M\cap N)$$
Edit:
This is a contradiction when $\Phi(N)$ is finitely generated because the Frattini subgroup of $N$ is the set of non-generators of $N$. That is $N=\langle \Phi(N),M\cap N\rangle$ implies that $N=M\cap N$ so $N\le M$. Hence $G=\phi(N)M\le NM=M$.
A: I've posted the question on MathOverflow and get a marvelous counterexample constructed by Ycor. 
A: As for whether $N$ needs to have the property that every proper subgroup is contained in some maximal subgroup of $N$. Wouldn't Klein-4 group $V_4=\langle a,b|a^2=b^2=1, ab=ba\rangle$ be a counterexample?
Take any subgroup $\langle a\rangle$, which is normal in $V_4$, and $\Phi(\langle a\rangle)=\langle a\rangle$ but $\Phi(V_4)=1$?
