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 $X'$.
Summarizing the proof defines $$ V = \left\{x : \left| \Lambda_i x \right| < r_i, 1 \leq i \leq n \right\} $$ If $\Lambda_1, \ldots, \Lambda_n \in X'$ and $r_i > 0$ for $i = 1,\ldots, n$.
$V$ is convex, balanced and $V \in \tau'$, the core of the proof is to prove that the collection of all $V$ forms a local base.
For the multiplication specifically we have
Suppose $x \in X$ and $\alpha$ is a scalar. Then $x \in sV$ for some $s>0$. If $\left|\beta - \alpha \right| < r$ and $y - x \in rV$ then $$ \beta y - \alpha x = (\beta - \alpha)y + \alpha(y - x) $$ lies in $V$ provided that $r$ is so small that $$ r(s + r) + \left| \alpha \right|r < 1 $$ Hence the scalar multiplication is continuous.
I'm missing here probably what the author is trying to prove exactly to show that the multiplication is continuous. Can you expound in detail what exactly is happening?
Update:
I suppose that for given $\alpha$ scalar and $x \in X$ the author wants to prove that for any $V$ such that $\alpha x \in V$ there's an open $U \times W$, $U$ neighborhood of $\alpha$ and $W$ neighborhood of $x$ such that if $\beta \in U$ and $y \in W$ we have $\beta y \in V$, but I'm confused why he proves that $\beta y - \alpha x \in V$ instead.
What Rudin is using is an equivalent definition of the notion of continuity for the multiplication. This definition together with the use of the balanced local base allow to prove the theorem.
In section 1.6 we have the following definition:
In theorem 3.10 It is proved first that the addition is continuous, so I'm going to use this fact as well.
I'll rephrase now the definition of continuity of the multiplication to get what Rudin is trying to prove
I believe is easy enough to prove that if $x \in X$ then there's an $s > 0$ such that $x \in sV$, where $V$ is a neighborhood of the form given in my question.
Now if $x \in X$, $\alpha$ scalar and $V$ is an open such that $\alpha x \in V$ then there's a neighborhood $V_0$ of $0$ such that $V = \alpha x + V_0$. If the multiplication is continuous we have $r > 0$ and a neighborhood $W$ of $x$ meeting the conditions mentioned earlier. This is equivalent to say there's a neighborhood $W_0$ of $0$ such that
$$ \beta (x + W_0) \subset \alpha x + V_0 $$
But this is equivalent to say
$$ \beta y \in \alpha x + V_0 \iff \beta y - \alpha x \in V_0 $$
and these need to hold for every $y \in x + W_0$, using the fact that the family $V$ forms a local you can derive
$$ y \in x + rV \iff y - x \in rV. $$
Conclusion is : to prove that the multiplication is continuous at $(\alpha,x)$ you need to prove that for every neighborhood $V$ of $\alpha x$ there's an $r > 0$ such that $\left|\beta - \alpha \right| < r$ and $y - x \in rV$ implies $\beta y - \alpha x \in V$