(1) If $x\neq 0$ is a vector in $X$, then there is a basis ${\cal B}$ of $X$ such that $x\in {\cal B}$ by the axiom of choice. Note that a linear functional $T:X\to \mathbb{R}$ is completely determined by its action on ${\cal B}$ (or, indeed, any basis of $X$).
(2) If $y\in {\cal B}$ but $y\neq x$, then we define $T(y)=0$; we also define $T(x)\neq 0$. Note that $T$ extends linearly to a linear functional on $X$. More precisely, if ${\cal B}=\{y_i\}_{i\in I}$ where $I$ is an index set and if $x=\Sigma_{i\in I} c_iy_i$ where all but finitely many coefficients in this sum are $0$, then we can define (and, in fact, are forced to define) $T(x)=\Sigma_{i\in I} c_iT(y_i)$.
Exercise 1: Let $V$ be a finite dimensional vector space and let $W$ be an arbitrary vector space. If $(v_1,\dots,v_n)$ is a basis of $V$ and if $(w_1,\dots,w_n)$ is an arbitrary tuple of vectors in $W$, prove that there exists a unique linear map $T:V\to W$ such that $T(v_i)=w_i$ for all $1\leq i\leq n$.
Exercise 2: Let $V$ be an arbitrary (real) vector space.
(a) Prove (using the axiom of choice) that $V$ has a basis ${\cal B}=\{y_i\}_{i\in I}$ (where $I$ is an index set).
(b) If $c_i\in \mathbb{R}$ for all $i\in I$, prove that there exists a unique linear functional $T:V\to \mathbb{R}$ such that $T(y_i)=c_i$ for all $i\in I$.
Challenge: Let $X$ be a (real) vector space. We say that $X$ is a normed linear space if there is a function $X\to [0,\infty)$ with the following properties (for notational convenience, we denote the image of $x\in X$ under this function by $\left\|x\right\|$):
- $\left\|x\right\|=0$ if and only if $x=0$.
- $\left\|\alpha\cdot x\right\|=\left|\alpha\right|\left\|x\right\|$ for all $x\in X$ and all scalars $\alpha\in \mathbb{R}$.
- $\left\|x+y\right\|\leq \left\|x\right\|+\left\|y\right\|$ for all $x,y\in X$.
A linear functional $g:X\to\mathbb{R}$ is said to be a bounded linear functional if
$\left\|g\right\|=\text{sup}\{\frac{\left|g(x)\right|}{\left\|x\right\|}:x\in X, x\neq 0\}<\infty$.
If $g:X\to\mathbb{R}$ is a bounded linear functional, then the norm of $g$ is defined to be the number $\left\|g\right\|$ (above).
Prove the Hahn-Banach theorem:
Let $X$ be a normed linear space (over $\mathbb{R}$, if you prefer) and let $Y\subseteq X$ be a subspace. If $f:Y\to \mathbb{R}$ is a bounded linear functional, prove that there exists a bounded linear functional $F:X\to\mathbb{R}$ such that $F(y)=f(y)$ for all $y\in Y$ and $\left\|F\right\|=\left\|f\right\|$.
I hope this helps!