Theorem 3.10 in Rudin's Functional Analysis 
Theorem 3.10: Suppose $X$ is a vector space and $X'$ is a separating vector space of linear functionals on $X$. Then the $X'$-topology $\tau'$ makes $X$ into a locally convex space whose dual space is $X'$.

In addition to having the same trouble as this questioner, I am also having trouble with this:

Conversely, suppose $\Lambda$ is a $\tau '$-continuous linear functional on $X$. Then $|\Lambda x| < 1$ for all $x$ in some set $V$ of the form (1). Condition (b) of Lemma 3.9 therefore holds; hence so does (a): $\Lambda = \sum \alpha_i \Lambda_i$.

(1) refers to $V = \{x : |\Lambda_i x| < r_i ~~\text{ for }~~ 1 \le i \le n \}$ and Lemma 3.9 says the following:

Lemma 3.9: Suppose $\Lambda_1,...,\Lambda_n$ and $\Lambda$ are linear functionals on a vector space $X$. Let $$N = \{x : \Lambda_1 x = ... = \Lambda_n x = 0\}.$$ The following three are then equivalent
(a) There are scalars $\alpha_1,...,\alpha_n$ such that $$\Lambda = \alpha_1 \Lambda_1 + ... + \alpha_n \Lambda_n$$
(b) There exists $\gamma < \infty$ such that $$|\Lambda x| \le \gamma \max_{1 \le i \le n} |\Lambda_i x| ~~~(x \in X)$$
(c) $\Lambda x = 0$ for every $x \in N$.

My question is, why is condition (b) satisfied?
 A: I complete the proof that was suggested by logarithm in the comment.
I start from
\begin{equation*}
V = \{x\in X:|\Lambda_ix|<r_i\text{ for }1\le i\le n\} \subset \{x\in X: |\Lambda x| < 1\}.
\end{equation*}
Put $\gamma = 2/\min\{r_1,\dots,r_n\}$ and $M_x = \max\{|\Lambda_1x|,\dots,|\Lambda_nx|\}$ for $x\in X$.
If $M_x = 0$ for all $x\in X$, then $V = X \subset \{x:|\Lambda x|<1\}$ implies that $|\Lambda x| = 0 = \gamma M_x$ for all $x\in X$.
Assume that $M_x > 0$ for some $x\in X$.
If $M_x > 0$, then $|\Lambda_i(x/(\gamma M_x))| = |\Lambda_ix|/(\gamma M_x) \le 1/\gamma < r_i$ for all $i$; thus $x/(\gamma M_x) \in V$ and $|\Lambda x| = \gamma M_x|\Lambda (x/(\gamma M_x))| < \gamma M_x$.
If $M_x = 0$, then there is $y\in X$ with $M_y >0$ by the assumption.
Let $0 = t_1 < t_2 < \cdots < 1$ with $t_j\to 1$ as $j\to\infty$ and $x_j = t_jx + (1-t_j)y$.
Since
\begin{align}
M_{x_j} &\ge |\Lambda_ix_j| = |t_j\Lambda_ix + (1-t_j)\Lambda_iy| \ge (1-t_j)|\Lambda_iy| \\
M_{x-x_j} &\ge |\Lambda_i(x-x_j)| = |(1-t_j)\Lambda_i(x-y)| \ge (1-t_j)|\Lambda_iy|
\end{align}
for all $i$ and $j$, $M_{x_j}\ge (1-t_j)M_y>0$ and $M_{x-x_j}\ge (1-t_j)M_y>0$ for all $j$.
Thus,
\begin{align*}
|\Lambda x| &\le |\Lambda(x-x_j)| + |\Lambda x_j| \le \gamma (M_{x-x_j} + M_{x_j}) \\
&\le \gamma [(1-t_j)(M_x + M_y) + t_j M_x + (1-t_j)M_y] \\
&= 2\gamma(1-t_j)M_y \to 0
\end{align*}
as $j\to\infty$.
Therefore, $|\Lambda x| = 0 = \gamma M_x$.
A: Here's an attempt to prove the equivalent condition (c).
More explicitly, I attempt to prove:
$$
N=\bigcap_{i}\ker\Lambda_i \subset \ker \Lambda
 $$
Let $V$ be a basic neighborhood of $0$  of the sort:
$$
V=\bigcap_{i} \{\ x:|\Lambda_i x|<r_i \ \} $$
Assume V satisfies: $$
|\Lambda x| \leq 1 \ \ \ :\ \ \  x\in V
$$
Obviously, $N\subset V$
Suppose there exist a $x\in N$ such that:
$$|\Lambda x|=c>0$$
Since $N$ is a subspace, for each $n\in \mathbb{N}$, we have:
$$
nx\in N \subset V
$$
Therefore:
$$
|\Lambda nx| = nc \underset{n\to \infty}{\to} \infty 
$$
This is absurd, since $\Lambda$ is bounded in $V$. This completes the proof of condition (c), which implies (b) and (a).
