Hahn Banach Theorem: Transfinite Induction In the 4th book of Stein and Shakarchi, the statement of the Hahn-Banach theorem is as follows:

Suppose $V_0$ is a linear subspace of $V$ and $p$ is a real sub-linear function on $V$, and that we are given a linear functional $l_0$ on $V_0$ that satisfies $$l_0(v)\leq p(v) ~\forall ~v\in V_0$$
  Then $l_0$ can be extended to a linear functional $l$ on $V$ that satisfies 
  $$l(v) \leq p(v)~ \forall ~ v \in V$$

The proof given is as follows:

We have already proved that if $l_0$ can be extended to a subspace W, then it can also be extended to the subspace $W+ w$ for any vector $w$. We well order all the vectors in $V$ that do not belong to $V_0$, and denote this ordering by $<$. Among these vectors, we call a vector $v$ extendable if the linear functional $l_0$ has an extension of the kind desired to the subspace spanned by $V_0.v$ and all the vectors $<v$. What we want to prove is in effect that all vectors not in $V_0$ are extendable. Assume the contrary, then because of the well-ordering we can find the smallest $v_1$ that is not extendable. Now if $V_0'$ is the space spanned by $V_0$ and all the vectors $<v_1$,then by assumption $l_0$ extends to $V_0'$. The previous step, with $V_0'$ in place of $V_0$ allows us then to extend $l_0$ to the subspace spanned by $V_0'$ and $v_1$, reaching a contradiction. This proves the theorem.

Firstly, how does proving that all vectors not in $V_0$ are extendable imply that an extension exists for the whole space? We know that for any $v \notin V_0$ there exists a valid extension, say $l_v$ for the space spanned by $V_0,v$ and all vectors $<v$. Say we define $l(v):=l_v(v)$. Now we take an arbitrary $w>v$. Couldn't it happen that $l_w(v) \ne l_v(v)$. In this case how can one define $l$? Will it help if we change the definition of "extendable" to something like: A vector $v \notin V_0$ is called extendable if $l_0$ has an extension of the kind desired to the relevant subspace which is equal to the already defined functionals $l_w~ \forall w<v$ in the relevant subspaces? Also, how does one get the statement :"Now if $V_0'$ is the space spanned by $V_0$ and all the vectors $<v_1$,then by assumption $l_0$ extends to $V_0'$." 
 A: I think your concerns about the proof are correct. Let me try to offer a repaired version of the proof. 
Well-order $V \setminus V_0$. Let $v_1$ be the least element of $V \setminus V_0$. Let $S_{\leq v}$ denote the span of $V_0 \cup \{ w: w \leq v \}$. 
Let $S_{< v}$ denote the span of $V_0 \cup \{ w: w < v \}$. 
We use (transfinite) recursion to define, for each $v \in V \setminus V_0$, a linear functional $\ell_v$ on $S_{\leq v}$ that is dominated by $p$, that extends $\ell_0$, and that extends $\ell_u$ for each $u < v$. It is then immediate that defining $\ell$ on $V$ by $\ell(v) = \ell_v(v)$ gives the desired functional (to verify linearity use that $\ell_v$ is extended by $\ell_w$ whenever $v<w$).  
For the basis step, define a linear functional $\ell_{v_1}$ on $S_{\leq v_1}$ that is dominated by $p$ and extends $\ell_0$ in the usual way. For the inductive step, suppose that for each $u < v$ we have defined a linear functional $\ell_u$ on $S_{\leq w}$ that is $p$-dominated and extends $\ell_0$. Suppose also that $\ell_t = \ell_u$ for each $t \leq u < v$. Define $\ell'_v$ on the span of $S_{< v}$ as follows. For each $u$ in $S_{<v}$, choose $u'<v$ such that $u \in S_{\leq u'}$ and define $\ell'_v(u) = \ell_{u'}(u)$. 
It is clear that $\ell'_v$ is dominated by $p$ and that $\ell'_v$ extends $\ell_u$ for each $u < v$. Using this last fact, it is easy to check that $\ell'_v$ is linear. Now extend $\ell'_v$ to a linear functional $\ell_v$ on $S_{\leq v}$ in the usual way. This completes the recursive definition. 
