Q:

Linearity of functionals imply the inclusion of kernels

I understand most of the following claim's proof except statement I've highlighted below:

Let $X$ be a linear space and $W$ be a subspace of $X^{\text{#}}$ (linear functionals - not necessarily bounded). Then a linear functional $\psi:X \rightarrow \mathbb{R}$ is $W$-weakly continuous if and only if it belongs to $W$.

Proof:

The reverse directions follows from definition. In the forward direction, suppose $\psi:X \rightarrow \mathbb{R}$ is $W$-weakly continuous. By the continuity of $\psi$ at $0$, there is a neighborhood $N$ of $0$ for which $|\psi(x)| = |\psi(x)-\psi(0)| < 1$ if $x \in N$. There is a neighborhood in the base for the $W$-weak toplogy (denoted below as $N_{\epsilon, \psi_1, \ldots, \psi_n}$) at $0$ that is contained in $N$. Choose $\epsilon > 0$ and $\psi_1,...\psi_n$ in $W$ for which:
  $$N_{\epsilon, \psi_1, \ldots, \psi_n}= \{x' \in X \,\big| \,\, |\psi_k(x')-\psi_k(0)| < \epsilon \quad 1 \leq k \leq n\} \subset N$$Thus:
  $$|\psi(x)|<1 \text{ if } |\psi_k(x)|<\epsilon \quad \forall 1\leq k \leq n$$
By linearity of $\psi$ and $\psi_k$ we have the inclusion $\bigcap \ker \psi_k \subset \ker \psi$. By a previous theorem, this implies $\psi$ is a linear combination of the $\psi_k$'s.

I don't understand the bolded statement. Namely, suppose $A \subset N_{\epsilon, \psi_1, \ldots, \psi_n}$ is the set of all $x$ for which all the $\psi_k$ are zero -- i.e. $A = \bigcap \ker \psi_k$. The implication shows $|\psi(A)|<1$. This doesn't imply $|\psi(A)|=0$. So how do we get $A \subset \ker \psi$?

A:

Assume that $x \in \bigcap_{k=1}^n \ker \psi_k$. Then $|\psi_k(x)| = 0 < \varepsilon$ for all $1 \le k \le n$ so it follows $|\psi(x)| < 1$.
But also $n x \in \bigcap_{k=1}^n \ker \psi_k$ for any $n \in \mathbb{N}$ so similarly $n|\psi(x)| = |\psi(nx)| < 1$. If $\psi(x) \ne 0$, the left hand side would be unbounded as $n \to \infty$ so it has to be $\psi(x) = 0$.

