Question about Nelson's proof of approximating real Banach space with a dense linear subspace equating with given linear functionals Below is a theorem from Nelson's Dynamical Theories of Brownian Motion. 
I have several questions regarding the proof. 
First, why does the general case follow from proving that if $\mathcal{D}$ is a dense convex subset and $\mathcal{M}$ is a closed affine hyperplane then $\mathcal{D} \cap \mathcal{M}$ is dense in $\mathcal{M}$? I can't see why this proves that for a fixed number of continuous linear functionals $u_1, \dots , u_n$ and $f \in \mathcal{X}$, $\delta > 0$, there is a $g \in \mathcal{D}$ with $\Vert f-g \Vert \le \delta$ and $u_1(f) = u_1(g) , \dots, u_n(f)=u_n(g)$. 
Second, why is the linear functional that assigns to each element of $\mathcal{X}$ the corresponding coefficient of $e$ continuous since $\mathcal{M}$ is closed? 
Third, I cannot see why $g = \frac{r_-g_+ + r_+ g_-}{r_- + r_+}=\frac{r_- m_+ + r_+ m_-}{r_-+r_+}$. Since $g_+ = m_+ + r_+ e$ and $g_- = m_- + r_- e$, shouldn't we have $g = \frac{r_- m_+ + r_+ m_-}{r_-+r_+}+ 2 \frac{r_- r_+}{r_- + r_+}e$?
Finally, why does $g$ converge to $f$ as $\epsilon \to 0$?


 A: Suppose the statement is true and note that $\mathscr M:=f+\ker(u_1)$ is an affine hyperplane. Then if $\mathscr D\cap \mathscr M$ is dense in $\mathscr M$ you can find a $g$ in it that has distance $<\delta$ to $f$. This $g$ then satisfies $u_1(g)=u_1(f)$, since $f-g\in\ker(u_1)$.
In the case of multiple linear functionals you proceed with the same argument. Look at $\mathscr M_n:= f + \bigcap_{k≤n}\ker(u_k)$, then if $\mathscr D\cap\mathscr M_n$ is dense you find a $g$ in it having distance $<\delta$ to $f$, by construction this $g$ satisfies $u_1(g)=u_1(f), ... u_n(g)=u_n(f)$. So why the intersection dense? Here you must perform an induction as remarked in your picture.
First note that $X_1:=\Bbb Rf + \ker(u_1)$ is either all of $X$ or a hyper-plane and as such $\mathscr D \cap X_1$ is dense in $X_1$, further it is Banach. Then $X_k:= \Bbb Rf +\ker(u_k)\cap X_{k-1}$ is either a hyper-plane or the same as $X_{k-1}$, as such $\mathscr D\cap X_k$ is dense in $X_k$ by induction. In the end this gives you that $\mathscr D\cap X_n$ is dense in $X_n$, which is Banach, and then it is dense in $f+\bigcap_{k≤n}(u_k)$, because this is now an affine hyper-plane of $X_n$.
For your second question, $\mathscr M$ is now the kernel of the linear functional that reads off the coefficient of $e$. If this linear functional were discontinuous then its kernel must be dense in $X$, but since $\mathscr M$ is closed this is not possible.
For your third question note that there is a typo in the image. You want $g_- = m_- - r_-e$, not $g_-= m_-+r_- e$. Without this definition the remark that $r_-\to1$ as $\epsilon\to0$ does not make any sense. With this definition the equality of the two expressions is obvious.
Finally note that both $m_+$ and $m_-$ converge to $f$ by definition and that $r_+, r_-$ converge to $1$. Hence $g=\frac{r_+m_++r_-m_-}{r_++r_-}$ converges to $\frac{f+f}{1+1}=f$.
