Let $E$ be a normed vector space and $\sigma_c(E',E)$ denote the topology of compact convergence on $E'$. It's easy to show that $\sigma_c(E',E)$ is Hausdorff.
Now let $\delta>0$ and $$V_\delta:=\{\varphi\in E':\left\|\varphi\right\|_{E'}\le\delta\}.$$ Clearly, the subspace topology $\left.\sigma_c(E',E)\right|_{V_\delta}$ is Hausdorff as well.
But how can we show that $V_\delta$ is $\sigma_c(E',E)$-compact? It is clearly $\left\|\;\cdot\;\right\|_{E'}$-bounded. Does this already imply the claim?
The answer is yes. The proof is contained essentially in the following fact:
If $E$ is a LCS and $A \subset E'$ is equicontinuous, then the topology of totally bounded convergence and the topology of pointwise convergence coincide on A.
Indeed, assume that this is true. Then $V_\delta$ is equicontinuous and compact in the topology of pointwise convergence due to Banach-Alaouglu theorem. Thus, it is compact in the topology of convergence on totally bounded sets. A fortiori it is compact in the topology of compact convergence.
To prove the foregoing fact fix $f \in A$ and a totally bounded set $K \subset E$. We are going to construct a finite set $F = \{x_1, \dots, x_n\} \subset E$ s.t. $(f + 1/3\;F^\circ) \cap A \subset (f + K^\circ) \cap A$. Let $U \subset E$ be a neighbourhood of $0$ s.t. $A \subset U^\circ$. Since $K$ is totally bounded there exists a finite set $F \subset E$ s.t. $F + 1/3 \; U \supset K$. Assume that $g \in (f + 1/3\; F^\circ) \cap A$. Then $g-f \in 1/3\; F^\circ$ and, therefore, $|(g-f)(x)| \le 1/3$ for all $x \in F$. Now for all $x \in K$ there exists $x' \in F$ s.t. $x-x' \in 1/3\; U$. Therefore $$|(g-f)(x)| \le |g(x) - g(x')| + |g(x') - f(x')| + |f(x') - f(x)| \le 1,$$ because $f,g \in U^\circ$ and $x-x' \in 1/3\;U$. Thus, $g \in (f + K^\circ) \cap A$.
Therefore, each neighbourhood of $f$ in $A$ in the topology of totally bounded convergence contains a smaller neighbourhood of $f$ in the topology of pointwise convergence. It easily follows that the topologies coincide.