I am working through the proof of the following proposition V.4.4.7 in Conway's functional analysis.
$\color{blue}{\text{If } X \text{ is a Banach space and } \Lambda \in X^*, \text{ then }\text{ker}(\Lambda) \text{ is proximal if and only if there is }x \in X, \|x\|=1,\text{ and }\Lambda x = \|\Lambda\|.}$
Now the beginning of the proof goes as follows:
$\implies$ Let $M := \text{ker}{\Lambda}$. Define $f: X/M \to \mathbb{F}$ via $f(x+M) = \Lambda x$, so clearly $f$ is a linear functional with $\|f\| = \|\Lambda\|$. Since $\text{dim}(X/M)=1$, there is $x \in X$ with $\|x+M\| = 1$ and $f(x+M) = \|f\|$. Etc.
The rest of the forward direction from there is very straightforward, but I do not understand this last sentence. It is basically saying that $\sup_{\|x\| = 1} f(x)$ is attained, but it is not obvious to me why a continuous linear map from technically "$\mathbb{R}$ to $\mathbb{R}$" would give us this attainment.
Similarly, in the backwards direction:
$\impliedby$ Let $x_0$ be the element such that the hypothesis holds. If $x \in X$ with $\|x+M\| = \alpha$, then $\|\alpha^{-1} x+ M\| = 1$. But it is also true that $\|x_0 + M\| = 1$ (I'm assuming the reason for this is the answer to my previous question). Then since $\text{dim}X/M = 1$, there is a $\beta \in \mathbb{F}$ with $|\beta|=1$ such that $\alpha^{-1} x+M = \beta(x_0 + M)$. Etc.
I once again am not sure why this last sentence is true. Maybe I am missing something obvious, but I'm not sure. Any help or hints to point me in the right direction would be much appreciated.
For the first point, since $\dim(X/M)=1<\infty$, the unit sphere of $X/M$ is compact. Thus taking a sequence $\{x_n+M\}$ in this sphere such that $f(x_n+M)\to\|f\|$, we have a subsequence with a limit point $x+M$, and by continuity one has $f(x+M)=\|f\|$.
For the second point, since $\dim(X/M)=1$, we have $X/M$ is isomorphic to $\mathbb F$. Let $g:X/M\to\mathbb F$ be such an isomorphism, and put $\beta=\frac{f(\alpha^{-1}x+M)}{f(x_0+M)}$. Then since $f((\alpha^{-1}x+M)-\beta(x_0+M))=0$, we must have $(\alpha^{-1}x+M)-\beta(x_0+M)=0$.