I am looking for verification of the following:
Let $X$ be a Banach space. We will show that if $M, N$ and $M+N$ are closed subspaces of $X$, then $$\dfrac{M+N}{N} \cong \dfrac{M}{M \cap N},$$ using the map $\phi : M \to \dfrac{M+N}{N}$, where $x \mapsto x + N$. For any $x,y \in M$, we have that, $$\phi(x+y) = (x+y) + N = (x + N) + (y + N) = \phi(x) + \phi(y)$$ Also, for any $\lambda \in \mathbb{F}$, we have, $$\phi(\lambda x) = (\lambda x) + N = \lambda(x + N) = \lambda\phi(x)$$ Then: \begin{align*} \ker(\phi) &= \left\{ m \in M : \phi(m) = e_{\frac{M+N}{N}} \right\} \hspace{1cm}\text{should this be sent to 0 instead of the "identity"?}\\ &= \left\{ m \in M : m + N = N \right\} \\ &= \left\{ m \in M : m \in N \right\} \\ &= M \cap N \end{align*} Then, we see that $\phi$ is a surjection because $$(m+n) + N = m + N \in \dfrac{M+N}{N}$$ which is $\phi(m)$. Then, we use the first isomorphism theorem, that is, for any operator $\phi: X \to Y$ between Banach spaces, $$X / \ker(\phi) \cong \text{im} \phi \iff \text{im}\phi \space\ \text{is closed in} \space\ Y$$ So that here, $X = M$, $\ker(\phi) = M \cap N$, and $Y = \text{im}(\phi) = \dfrac{M+N}{N}$, so that the result follows.
Indeed, your proof should mirror that of the second isomorphism theorem. In fact, this statement is (almost) an instance of the second isomorphism theorem for modules, except for the fact that our $\phi$ (and eventually our isomoprhism) must also be continuous.
Your proof is correct except for two things: