I'm doing this exercise. In my proof, I use both closed graph theorem and open mapping theorem. Could you have a check if I made logical mistakes?
Let $(E, |\cdot |_E)$ be a Banach space and $M$ a closed subspace of $E$. Consider the quotient map $$T: E \to E/M, x \mapsto \hat x := x+M.$$ Then $T$ is linear and surjective. We endow $E/M$ is with the quotient norm $\| \cdot \|$ defined by $$\| \hat x\| := d(x, M), \quad x\in E.$$ Then $\| \cdot \|$ induces the quotient topology on $E/M$.
My proof:
First, let's prove that $\operatorname{graph} T := \{(x, \hat x) \mid x\in E\}$ is closed in $E \times (E/M)$. Let $(x_n, \hat x_n)$ be a sequence in $\operatorname{graph} T$ such that $(x_n, \hat x_n) \to (x, \hat a)$ for some $x,a \in E$. Then $\|\hat x_n – \hat a\| \to 0$ and thus there is a sequence $(z_n)$ in $M$ such that $|x_n-a-z_n|_E < 1/n$. Because $x_n \to x$, we get $z_n \to x-a$. Because $M$ is closed, $x-a\in M$. This implies $\hat a = \hat x$ and thus $(x, \hat a)=(x, \hat x) \in \operatorname{graph} T$. As such, $\operatorname{graph} T$ is closed.
Let $U$ be a subset of $E/M$ such that $V :=T^{-1} [U]$ is open in $E$. We want to prove that $U$ is open in the norm topology of $E/M$. Notice that $T$ is surjective and thus $T[V] = U$, so it suffices to show that $T$ is open. Indeed, $T$ is continuous by closed graph theorem. By open mapping theorem, $T$ is open. This completes the proof.
Update: This answer obtains the continuity of $T$ in a beautiful way as follows. We have $$\|Tx\| = \|\hat x\| = d(x, M) := \inf_{y\in M} |x-y|_E \le |x-0|_E = |x|_E.$$ Hence $T$ is continuous at $0$ and thus continuous.
Best Answer
You might want to justify the existence of the $z_n \in M$ better? It's true but you should be explicit.
A more serious problem is that you use the closed graph theorem and open mapping theorem before checking that $X{/}M$ is a Banach space (completeness is essential in these theorems, see Wikipedia, e.g.) Or has that been checked in your text and is it a quotable theorem? Then it's OK.
The answer you linked also showed directly and rather simply that $T$ is open by showing $T[B_E(0,1)]= B_{X{/}M}(T(0), 1)$ (linearity does the rest). Why not use that instead of getting out the heavy guns?