You are correct and the proof looks sufficiently rigorous. However, $\|T\|_\text{op}$ can be computed exactly. That is, $\|T\|_\text{op}=\dfrac12$. To show this, let $z=(z_1,z_2,z_3,\ldots)\in\ell^\infty$. Then,
$$T(z)=\left(\frac{z_2}{2},\frac{z_3}{3},\frac{z_4}{4},\ldots\right)$$
so that
$$\big\|T(z)\big\|_{\infty}=\sup\left\{\frac{|z_k|}{k}\,\Big|\,k=2,3,4,\ldots\right\}\leq \sup\left\{\frac{\|z\|_\infty}{k}\,\Big|\,k=2,3,4,\ldots\right\}=\frac{\|z\|_\infty}{2}\,.$$
Note that the equality holds for $z=(0,1,0,0,0,\ldots)$. This implies $\|T\|_{\text{op}}= \dfrac{1}{2}$.
You can write an explicit solution $x\in\ell^\infty$ to $x=y+T(x)$. That is, $$x=(1-T)^{-1}y=\left(\sum_{k=1}^\infty\,\frac{y_k}{k!},\sum_{k=2}^\infty\,\frac{2!y_k}{k!},\sum_{k=3}^\infty\,\frac{3!y_k}{k!},\ldots\right)$$
Nonetheless, you did sufficient and good work. I was just making additional comments.
Let $x \in \mathbb{R}$. For all $n \geq 1$, apply the hypothesis on the interval $[x, x + \frac{1}{n}]$ : there exists $x_n$ in this interval such that
$$f(x_n)=g(x_n) \quad \quad (1)$$
Now because $x \leq x_n \leq x + \frac{1}{n}$, you have that $(x_n)$ tends to $x$. Let $n$ tend to $+\infty$ in the equality $(1)$ : by continuity of $f$ and $g$, you get
$$f(x)=g(x)$$
Best Answer
Long comment.
(I). If $A$ is a vector subspace of the normed linear space $X$ and $f$ is a linear functional on $A$ with $\sup \{|f(a):\|a\|=1\}=1$ then $f$ extends uniquely to a functional $\bar f$ on $\bar A$ such that $\sup \{|\bar f(\bar a)|: \|\bar a\|=1\}=1: $ Show that if $(a_n)_n$ and $(a'_n)_n$ are any sequences in $A$ converging to $\bar a\in \bar A,$ then $(f(a_n))_n$ and $(f(a'_n))_n$ converge, and converge to the same value, which we define as $\bar f(\bar a).$ We can then extend $\bar f$ to $f'\in X^*$ with $\|f'\|=1$ by Hahn-Banach applied to $\bar f$.
(II). In the Q, any $y=(y_n)_{n\in \Bbb N}$ is in $M$ iff $y$ has bounded partial sums, i.e. $\infty>\sup \{|\sum_{j=1}^ny_j|: n\in \Bbb N\}.$ As has been shown in the A by Yanko, $M$ is not closed.
(III). A variant of the proof of 2.53 is to take $M'$ to be the vector sub-space of those members of $l_{\infty}$ that have Cesaro Means, and for $x=(x_n)_{n\in \Bbb N}\in M'$ let $L(x)=\lim_{n\to \infty}\frac {1}{n}\sum_{j=1}^n x_j.$ And extend $L$ to a member of $l^*_{\infty}$ with $\|L\|=1.$
(IV). There are uncountably many $L$ satisfying 2.53. They are called generalized limits.
In particular, there exists a binary sequence $y\in l_{\infty}$ \ $\overline {M'}$ with $\|y\|=d(y,\overline {M'})=1,$ and if you look at the details of the proof of the Hahn-Banach theorem, you will see there are uncountably many consistent possible choices for $L(y)$ when extending $L$ from $\overline {M'}$ to the vector space generated by $\{y\}\cup \overline {M'}.$