I'm trying to prove this lemma which then is used to prove this result. I have found another proof here which is much more complex then mine. Could you have a check on my proof?
Let $(E, |\cdot|)$ be a Banach space and $M$ a closed subspace of $E$. Consider the quotient map $$T: E \to E/M, x \mapsto \hat x := x+M.$$ We endow $E/M$ with the quotient norm $\| \cdot \|$ defined by $$\| \hat x\| := d(x, M), \quad \forall x\in E.$$ Prove that $(E/M, \| \cdot \|)$ is a Banach space.
My attempt: Clearly, $T$ is linear surjective. We have $$\|Tx\| = d(x, M) := \inf_{y\in M} |x-y| \le |x-0| = |x|.$$ Hence $T$ is continuous at $0$ and thus continuous. Let $(\hat x_n)$ be a Cauchy sequence in $E/M$, i.e., $$\forall \epsilon>0, \exists N \in \mathbb N, \forall m,n \ge N: d(x_m-x_n, M) < \epsilon.$$
By axiom of choice, there is a subsequence $(x_{\varphi(n)})$ such that $$d(x_{\varphi(n)} - x_{\varphi(n-1)}, M) < 2^{-n}, \quad \forall n \in \mathbb N.$$
Let $y_n := x_{\varphi(n)}$ and $z_0 :=0$. There is $z_1 \in M$ such that $|(y_1-z_1) - (y_0-z_0)| < 2^{-1}$. Similarly, there is $z_2 \in M$ such that $|(y_2-z_2) - (y_1-z_1)| < 2^{-2}$. Recursively, there is $z_n \in M$ such that $$|(y_n-z_n) - (y_{n-1}-z_{n-1})| < 2^{-n}.$$
Let $t_n := y_n - z_n$. Then $(t_n)$ is a Cauchy sequence in $E$. Because $E$ is complete, there exists $t \in E$ such that $t_n \to t$. Because $(\hat y_n)$ is a subsequence of the Cauchy sequence $(\hat x_n)$, it suffices to show that $\hat y_n \to \hat t$. Notice that $y_n - t_n = z_n \in M$, so $\hat y_n = \hat t_n$. Hence it suffices to show that $\hat t_n \to \hat t$. In fact, $\hat t_n \to \hat t$ because $T$ is continuous and $t_n \to t$. This completes the proof.
