Here is the Proof prove of the existence of unbounded linear functionals in an infinite dimensional normed space $X.$
Example 4.2. Let $X$ be an infinite-dimensional Banach space, and let $\left\{x_{i}\right\}_{i \in I}$ be a Hamel basis for $X$. By dividing each vector by its norm, we can assume that $\left\|x_{i}\right\|=1$ for every $i \in I .$ Let $J_{0}=\left\{j_{1}, j_{2}, \ldots\right\}$ be any countable subsequence of $I .$ Define a scalar-valued function $\mu$ on $\left\{x_{i}\right\}_{i \in I}$ by setting $\mu\left(x_{j_{n}}\right)=n$ for $n \in \mathbf{N}$ and $\mu\left(x_{i}\right)=0$ for $i \in I \backslash J_{0} .$ Then extend $\mu$ linearly to all of $X:$ Each nonzero vector $x \in X$ has a unique representation as $x=\sum_{k=1}^{N} c_{k} x_{i_{k}}$ for some $i_{1}, \ldots, i_{N} \in I$ and nonzero scalars $c_{1}, \ldots, c_{N},$ so we define $\mu(x)=\sum_{k=1}^{N} c_{k} \mu\left(x_{i_{k}}\right) .$ We also set $\mu(0)=0 .$ Then $\mu$ is a linear functional on $X,$ but since $\left\|x_{j_{n}}\right\|=1$ yet $\left|\mu\left(x_{j_{n}}\right)\right|=n,$ the functional $\mu$ is unbounded. $\quad \diamond$
There is a typo, the extension should be called $\tilde{\mu}.$
Why is $\mu$ at the last line unbounded?
Firstly, one can't say there's a typo with absolute certainty. More likely than not, it's an abuse of notation (one that you will frequently see).
A linear map $T:X\to Y$ between two normed spaces is said to be bounded if there exists an $M>0$ such that for all $x\in X$ with $\|x\|\leq1$, we have $\|T(x)\|\leq M$. That is, $T$ is bounded if the map $f_T:\{x\in X:\|x\|\leq 1\}\to\mathbb R$ given by $f_T(x)=\|T(x)\|$ is bounded
To show that the map $\mu$ constructed is not bounded, fix some real number $M>0$. It suffices to find some $x\in X$ with $\|x\|\leq 1$ with $|\mu(x)|>M$. To this end, if $n\in\mathbb N$ and $n>M$, then $$|\mu(x_{j_n})|=n>M,$$ and therefore $\mu$ is not bounded.