Vector in kernel of all linear functionals must be $0$ 
Let $E$ be a vector space over a field $K$. Suppose that $e\in E$ is such that $\forall f\in \mathcal L(E,K)$, $f(e)=0$. Then prove that $e=0$.

Elements of $\mathcal L(E,K)$ need not be continuous. This is claimed in the Remark following Definition 3.2 in Fabian's Banach Space Theory. I quote "[it] follows from a simple linear algebra argument".
What is this "simple linear algebra argument" they are alluding to ?
With the axiom of choice, one could define a basis $(b_i)_{i\in I}$ of $E$ and the corresponding coordinate functionals $f_i$. Then obviously $e=0$.
Is there a simpler argument, preferably something that doesn't resort to choice ?
 A: Some choice is needed, since it is consistent with $\mathsf{ZF}$ to have a nontrivial vector space $V$ such that $V^\ast=\{0\}$, and I'm talking about the algebraic dual here.
With choice given $e\in V$ with $e\neq 0$ it's easy to come up with a functional which is not zero on $e$: extend $\{e\}$ to a basis $\{e\}\cup\{v_i\mid i<\kappa\}$ of $V$ and define $\phi:V\to\Bbb R$ by $\phi(e)=1$ and $\phi(v_i)=0$ for every $i$. Extend by linearity to the whole space.
Note that in the context of Banach spaces we can even come up with a continuous functional which is nonzero on $e$, by using Hahn-Banach to extend the functional $\phi\colon\langle e\rangle\to\Bbb R$ given by $\phi(ae)=a$ to a bounded functional on the whole space.
A: Follows what might be called a "simple linear algebra argument" provided we allow its use of choice (in the form of Zorn's lemma) is "simple"; in any event, I don't see how "choice $\equiv$ Zorn's" can be avoided here.
If we can prove that, for any
$0 \ne e \in E \tag 1$
there exists a subspace
$V \subset E \tag 2$
such that
$E = \langle e \rangle \oplus V, \tag 3$
then we may define a functional
$f:E \in \mathscr L (E, K)  \tag 4$
via
$f(e) = \alpha \in K, \; \alpha \ne 0, \tag 5$
$f(ke) = k \alpha, \; k \in K, \tag 6$
$f(v) = 0, \; v \in V; \tag 7$
we note that (5) affirms that
$f(e) \ne 0;  \tag 8$
thus, as long as (1)-(3) bind, we may assert that 
$\exists f \in \mathscr L(E, K), \; f(e) \ne 0.  \tag 9$
We proceed to prove the existence of a subspace $V$ satisfying (3).  Let $\mathscr  V$ be the collection of subspaces of $E$ such that
$W \in \mathscr V \Longleftrightarrow e \notin W; \tag{10}$
$\mathscr V$ is clearly partially ordered by set inclusion, and if we have a chain
$\mathscr C \subset \mathscr V, \tag{11}$
that is, 
$X, Y \in \mathscr C \Longrightarrow X \subset Y \; \text{or} \; Y \subset X, \tag{12}$
then
$\displaystyle \bigcup_{W \in \mathscr C} W \tag{13}$
is clearly an upper bound for $\mathscr C$; thus by Zorn's lemma there is a maximal element $V \in \mathscr V$; I claim such a $V$ satisfies (3).  For if not, 
$\langle e \rangle \oplus V \subsetneq E, \tag{14}$
so 
$\exists \eta \in E, \eta \notin \langle e \rangle \oplus V; \tag{15}$
then
$\eta \notin V, \tag{16}$
and hence
$V \subsetneq \langle \eta \rangle \oplus V; \tag{17}$
furthermore
$e \notin \langle \eta \rangle \oplus V, \tag{18}$
lest
$e = s\eta + v, \; s \in K, v \in V \tag{19}$
with
$s \ne 0 \tag{20}$
lest 
$e \in V; \tag{21}$
in light of (19) and (20) we may write
$\eta = s^{-1}(e - v) \in \langle e \rangle \oplus V, \tag{22}$
which contradicts (15); hence (17) and (18) both hold and together they imply that $V$ is not maximal in the set $\mathscr V$ of subspaces not containing $e$, contradicting its construction via Zorn's; this contradictions in fact implies that $V$ satisfies (3), and hence (9) binds as well; thus we must have
$e = 0 \tag{23}$ 
if
$\forall f \in \mathscr L(E, K), \; f(e) = 0. \tag{24}$
$OE\Delta$.
