The following is a proposition from Folland:
Suppose $X$ and $Y$ are vector spaces with topologies defined, respectively, by the families $\{p_\alpha\}_{\alpha \in A}$ and $\{q_\beta\}_{\beta \in B}$ of seminorms, and $T: X \rightarrow Y$ is a linear map. Then $T$ is continuous iff for each $\beta \in B$ there exist $\alpha_1,\ldots, \alpha_k \in A$ and $C > 0$ such that $q_\beta(Tx) \leq C \sum_1^k p_{\alpha_j}(x)$.
The proof for one direction is obvious and follows from linearity. It is the converse that is causing me some trouble in following Folland's proof. I've included his proof below in chunks, interrupting in sections where I am unclear.
If $T$ is continuous, for every $\beta \in B$ there is a neighborhood of 0 in $X$ such that $q_\beta(Tx) < 1$ for $x \in u$.
How does continuity ensure such a ball of radius $1$ and centered at $0$? Is it because the open sets of $Y$ are all balls with radius less than $r$ for some $r \geq 0$?
By Theorem 5.14a we may assume that $U = \bigcap_1^k U_{0\alpha_j\epsilon_j}$, where $$U_{x\alpha\epsilon} = \{y \in X : p_\alpha(y-x) < \epsilon\}.$$ Let $\epsilon = \min(\epsilon_1, \ldots, \epsilon_k)$; then $q_\beta(Tx) < 1$ whenever $p_{\alpha_j}(x) < \epsilon$ for all $j$.
Now, given $x \in X$ there are two possibilities. If $p_{\alpha_j}(x) > 0$ for some $j$, let $y = \epsilon x\big/\sum_1^k p_{\alpha_j}(x)$. Then $p_{\alpha_j}(y) < \epsilon$ for all $j$. So, $$q_\beta(Tx) = \sum_1^k \epsilon^{-1} p_{\alpha_j}(x)q_\beta(Ty) \leq \epsilon^{-1}\sum_1^k p_{\alpha_j}(x).$$
I fail to see the motivation behind this definition of $y$, and how Folland concludes that $p_{\alpha_j}(y) < \epsilon$ for all $j$. I also am having trouble following the last equation above.
On the other hand, if $p_{\alpha_j}(x) = 0$ for all $j$, then $p_{\alpha_j}(rx) = 0$ for all $j$ and all $r > 0$, hence $rq_\beta(Tx) = q_\beta(T(rx)) < 1$ for all $r > 0$, hence $q_\beta(Tx) = 0$. Thus $q_\beta(Tx) \leq \epsilon^{-1}\sum_1^k p_{\alpha_j}(x)$ in this case too, and we are done.
This part of the proof is clear, but I have added it for completeness. It is mainly the case where $p_{\alpha_j}(x) > 0$ that is confusing me.
If you have topogies given by the seminorms (especially uncountable) don't think of a metric. These are seminorms that define topology, not metric.
Use the following definition of continuity: $T$ is continuous iff
Here the set $\{y\in Y: q_\beta(y)<1\}$ is used.
The idea of $y$ is the following. We would like to estimate $q_\beta(Tx)$. We obtained an estimate: $q_\beta(Tx)<1$ but only if $x$ is small, i.e. if $p_{\alpha_j}(x)<\varepsilon$ for all $j$. But our $x$ is any. Therefore we scale the vector $x$, consider $y=tx$ for sufficiently small $t$. If $p_{\alpha_j}(tx)<\varepsilon$ for all $j$ then $q_\beta(T(tx))<1$, that is $q_\beta(Tx)<1/t$. This is our desired estimate. Now we need to find the proper $t$. $$p_{\alpha_j}(tx)<\varepsilon \iff t<\frac{\varepsilon}{p_{\alpha_j}(x)}.\tag{1}$$ And this must be satisfied for all $j$. The easiest way is to define $$t=\frac{\varepsilon}{2\sum_j p_{\alpha_j}(x)}.$$ The factor $2$ is needed to guarantee (1) in case if only one $p_{\alpha_j}(x)>0$ (to avoid equality in (1)). This shows that there's a tiny mistake in your proof.
The above calculations show how to conclude that $p_{\alpha_j}(y)<\varepsilon$, but we can do it directly, without above explanation of the idea behind the choice of $y$. Namely, using the definition without the factor $2$: $$p_{\alpha_j}(y) = p_{\alpha_j}\left(\frac{\varepsilon x}{\sum_i p_{\alpha_i}(x)}\right) = \frac{\varepsilon }{\sum_i p_{\alpha_i}(x)} p_{\alpha_j}(x)\leq \varepsilon.$$
The last problematic estimate is just $$q_\beta(Tx) = \frac 1t q_\beta(T(tx))<\frac 1t.$$