Can any vector not in a linear space be separated from it? For context, I was exploring a bit more this question I asked before.
Suppose that $E$ is a vector space such that its algebraic dual $E^\ast$ can separate point, i.e. for all $x\in E\setminus\{0\}$, there is $f\in E^\ast$ such that $f(x)\neq 0$.
I am wondering if the following can be proved or disproved :

If $S$ is a linear subspace of $E$, and $x\notin S$, then there is $f\in E^\ast$ such that $f(x)\notin f(S)$

I am not sure exactly on how to address this, but first it is obvious that such a $f$ will be such that $f(S)=\{ 0\}$, otherwise $f(S)=\mathbb R$. Therefore we are looking for $f\in E^\ast$ such that $f(x)\neq 0$ and $f(S)=\{0\}$. Otherwise, I thought about assuming the statement is wrong and then try to create a point $y\in S$ such that $f(x)=f(y)$ for all $f\in E^\ast$ which would prove that $x=y\in S$, maybe some sort of projection of $x$ on $S$ but I don't know how to do that from just $E^\ast$ and no norm. Any input would be much appreciated.
 A: The following result needs Zorn's lemma ($\Leftrightarrow$ axiom of choice)
Let $V\subseteq E$ be a subspace and $\mathcal M = \{W\subseteq E : W \text{ is subspace and } W\cap V = \{0\}\}$. We  give $M$ a partial order via $V_1 \leq V_2 :\Leftrightarrow V_1 \subseteq V_2$.
Let $\Gamma \subseteq \mathcal M$ be a chain (a totally ordered subset). We show that $\Gamma$ has an upper bound. Note that $\{0\}\in \mathcal M \neq \emptyset$.
Case $\Gamma = \emptyset$: Any element $x\in\mathcal M$ is an upper bound for $\Gamma$.
Case $\Gamma \neq \emptyset$: Define
\begin{align*}
R = \bigcup_{S\in\Gamma} S
\end{align*}
Now $R\cap V = \{0\}$ and $R$ is a subspace of $E$, as one verifies as follows: Let $x_1, x_2\in R$. Then $\exists S_1, S_2 \in \Gamma$ with $x_1\in S_1, x_2\in S_2$. Since $\Gamma$ is totally ordered, we may assume (by possibly interchanging $x_1$ and $x_2$) that $S_1 \leq S_2$, so $x_1,x_2 \in S_2$. Thus for $s,t\in\Bbb K$ we have $sx_1+tx_2 \in S_2 \subseteq R$. Thus $R\in\mathcal M$ and $R$ is an upper bound for $\Gamma$, since $S\subseteq R$ for every $S\in \Gamma$.
Hence every chain in $\mathcal M$ has an upper bound in $\mathcal M$ and we imploy Zorn's lemma to get a maximal element $W\in\mathcal M$. By definition $W\cap V = \{0\}$.
It remains to show $V+W = E$. Suppose $x_0 \in E\setminus (V+W)$.
We show $(W+\Bbb K x_0) \cap V = \{0\}$. If $w+tx_0 \in (W+\Bbb K x_0)\cap V$, then $tx_0 \in -w+V \subseteq W+V$, thus $t=0$ (as $x_0\neq 0$). So $w=w+tx_0 \in W\cap V =\{0\}$, so $w=0$.
Thus $(W+\Bbb K x_0) \cap V = \{0\}$ and $W\subsetneq W+\Bbb K x_0$, so $W$ would not be maximal in $\mathcal M$ $\unicode{x21af}$.
Thus $E=V\oplus W$.

To get your desired result we take $V=\Bbb K x$. Now $E=\Bbb K x \oplus W$ for some $W\subseteq E$. Let
$$
\lambda\colon E=\Bbb K x \oplus W \mapsto \Bbb K,\ tx+w \mapsto t.
$$
Then $\lambda\in E^*$and $\lambda(x) = 1 \not\ni \{0\} = \lambda(W) \supseteq \lambda(S)$.
