Linear functional $\psi:X \to \mathbb{R}$ is $W$-weakly continuous if and only if it belongs to $W$ I'm trying to understand the proof of the following proposition from Royden and Fitzpatrick (Chapter 14) - the step that I highlight is in which I would like confirmation on my justification. I provide the full proof for completeness.

Proposition 5: Let $X$ be a linear space and W a subspace of $X^\#$ (space of linear functionals on X). Then a linear functional $\psi: X \rightarrow \mathbb{R}$ is W-weakly continuous if and only if it belongs to W.

Definition of "$\mathcal{F}$-weakly continuous":
If $\mathcal{F}$ is any collection of real-valued functions on a set $X$, the $\mathcal{F}$-weak topology is defined to be the weakest topology on $X$ for which all the functions in $\mathcal{F}$ are continuous. A function that is continuous with respect to the $\mathcal{F}$-weak topology is called $\mathcal{F}$-weakly continuous.
Proof:
By definition of the W-weak topology, each linear functional in $W$ is $W$-weakly continuous. It remains to prove the converse. Suppose the linear functional $\psi: X \rightarrow \mathbb{R}$ is $W$-weakly continuous . By the continuity of $\psi$ at $0$, there is a neighborhood $\mathcal{N}$ of $0$ for which $|\psi(x)| = |\psi(x)- \psi(0)| < 1$ if $x \in \mathcal{N}$. There is a neighborhood in the base for the $W$-topology at $0$ that is contained in $\mathcal{N}$. Choose $\epsilon > 0$ and $\psi_1, ...,\psi_n$ in $W$ for which $\mathcal{N}_{\epsilon,\psi_1,...,\psi_n} \subset \mathcal{N}$. Thus 
$$ |\psi(x)| < 1 \text{ if } |\psi_k(x)|<\epsilon \text{ for all } 1\le k \le n$$

By the linearity of $\psi$, and $\psi_k$'s, we have the inclusion $\bigcap_{k=1}^n \ker \psi_k \subset \ker \psi$.

By Proposition 4, $\psi$ is a linear combination of $\psi_k$. Therefore, since W is a linear subspace, $\psi \in W$. $$\tag*{$\Box$}$$
How I understand this is as follows: The step I highlight implies $\bigcap_{k=1}^n \{x: \psi_k(x) < \epsilon\} \subset \{x: \psi(x) < 1\}$. The $\epsilon$ has been fixed, and the $\psi_k$ rely on said epsilon. Is the idea that by linearity, for any $c \in \mathbb{R}$,
\begin{align*}
\bigcap_{k=1}^n \{x: \psi_k(x) < \epsilon\} \subset \{x: \psi(x) < 1\}  \Leftrightarrow& \bigcap_{k=1}^n \{x: \psi_k(cx) < \epsilon\} \subset \{x: \psi(cx) < 1\}\\ \Leftrightarrow& \bigcap_{k=1}^n \{x: \psi_k(x) < \frac{\epsilon}{c}\} \subset \{x: \psi(x) < \frac{1}{c}\} 
\end{align*}
Taking c arbitrarily large will yield $\bigcap_{k=1}^n \ker \psi_k \subset \ker \psi$ ?
 A: You have the right kind of idea but I think that it is somewhat unclear from your argument why taking $c$ arbitrarily large will give the desired result (since this involves taking a limit of sets). One way to justify this is to conclude from what you've written that
\begin{align*}
\bigcap_{c \in \mathbb{N}} \bigg(\bigcap_{k=1}^n \{x: \psi_k(x) < \frac{\epsilon}{c}\}\bigg) \subset \bigcap_{c \in \mathbb{N}} \bigg(\{x: \psi(x) < \frac{1}{c}\} \bigg)
\end{align*}
and then convince yourself that the left hand side is $\bigcap_{k=1}^n \ker \psi_k$ and the right hand side is $\ker \psi$. If I'm honest, this argument feels unnecessarily complicated to me.
A much simpler way to do this kind of argument (that avoids set builder notations entirely) is to just take $x \in \bigcap_{k=1}^n \ker \psi_k$ and show it's in $\ker \psi$, using essentially the same idea as is used in your argument. Indeed, since $0 = |\psi_k(cx)| < \epsilon$ for all $c > 0$, $|\psi(cx)| < 1$ and so $|\psi(x)| < \frac1c$. Since $c>0$ was arbitrary, $|\psi(x)| = 0$ so $x \in \ker \psi$. 
A: Your argument is correct. But (perhaps) a better argument is the following: let $\psi_k(x)=0$ for $1\leq k \leq n$. Let $n$ be any positive integer. Then  $\psi_k(Nx)=N\psi_k(x)=0$ for $1\leq k \leq n$. Hence $\psi (Nx) <1$ and $\psi (x) <\frac 1 N$. Since this holds for all $N$ we get $\psi (x) \leq 0$. Changing $x$ to $-x$ we get $\psi (x)\geq 0$. Hence $\psi (x)=0$.
