On Hahn-Banach Theorem The following is the first part of a proof for Hahn-Banach Theorem (Extension of linear functionals) from Kreyszig's book of Functional Analysis:

I don't undertsand the blue-underlined sentence of the text above. 
My questions are:
1- How each $D(g)$ is a vector space? Suppose $x_1, \ x_2 \in D(g)$ then $g(x_1) \le p(x_1)$ and $g(x_2) \le p(x_2)$. Then $g(x_1+x_2) = g(x_1)+g(x_2) \le p(x_1)+p(x_2)$ does not imply $g(x_1+x_2) \le p(x_1+x_2)$, because we have $p(x_1+x_2)\le p(x_1+x_2)$ by definition. So How $x_1+x_2 \in D(g)$?! The book has considered $a \ge 0$, so $g(ax) = ag(x) \ge ap(x)$. So the problem is just the sum inequality. 
2- How $\bigcup D(g)$ is a vector space because "$C$ is a chain"? I can't see a coonection.      
 A: Although it is not said explicitly in the proof, each $\mathcal D (g) $ is a subspace. And an
increasing union of subspaces is a subspace (this is where you use that $C $ is a chain).
As for your argument with  $a<0 $, try it for instance on $X=\mathbb R^2$, $f (x,y)=x+y $, $p (x,y)=|x|+|y|$.
A: Take $x,y \in D(k)=\bigcup\limits_{g\in C} D(g)$. Then $x \in D(g_1)$ and $y\in D(g_2)$ for some $g_1,g_2 \in C$. Now note that since $C$ is a chain we must have $g_1\le g_2$ or $g_2\le g_1$. Without loss of generality, assume $g_1\le g_2$. Then by definition $D(g_2) \supset D(g_1)$. Thus $x,y \in D(g_2)$. Now since $g_2 \in E$, we have
$$x+ay \in D(g_2) \subset D(g), \ \ \ \ k(x+ay)=g_2(x+ay)\le p(x+ay)$$
A: The thing is, you must understand that a Linear Extension of a functional $f:V\rightarrow\mathbb{R}$ is another functional $g:U\rightarrow\mathbb{R}$ such that


*

*$V\subset U$

*$g(x)=f(x)$, if $x\in V$


And functionals are defined in Linear subspaces.
The important thing in the proof is showing that the set $E$ is not empty, and we know it's not because $f\in E$. The other elements of $E$ will be functionals that extend $f$, i.e functionals $g:D(g)\rightarrow\mathbb{R}$, such that $Z\subset D(g)$ and that satisfy the two conditions in the text, by definition $D(g)$ are linear subspaces, don't worry too much about this, for the means of the proof the important thing is that $E\neq \emptyset$, it might be the case that $f$ is the only element in $E$, it doesn't matter.
The others already answered your question about $\bigcup D(g)$ being a linear space.
And if the inequality you stated doesn't hold for any $x\in D(g)$ then it's just the case that $g\not\in E$, but again, the only important thing is that $E\neq\emptyset$.
