Suppose that $X$ is a metric space and let $C_k(X)$ denote the space of real functions on $X$ with the topology of uniform convergence on compact sets. Then $C_k(X)$ is a topological vector space. Let $C_k(X)^*$ denote the strong dual space. For each $x\in X$ the point evaluation functionals $\delta_x$ belong to $C_k(X)^*$.
Is the set $\{\delta_x\colon x\in X\}$ bounded in $C_k(X)^*$?
When $X$ is compact this is trivial.
$S \subseteq C_k(X)^*$ is bounded in the strong dual topology if and only if $\sup _{\omega \in S, f \in B} |\omega (f)| < +\infty$ for every bounded $B \subseteq C_k(X)$.
$B \subseteq C_k (X)$ is bounded if and only if $\sup _{f \in B, x \in K} |f(x)| < +\infty$ for every $K \subseteq X$ compact.
Call your subset $S$. Assume that $X$ is not compact; for metric spaces, compactness is equivalent to pseudocompactness, therefore there exists an unbounded $F \in C_k(X)$; notice that $\{F\} \subset C_k(X)$ is obviously bounded. Then
$$\sup _{\omega \in S, f \in \{F\}} |\omega(f)| = \sup _{x \in X} |F(x)| = +\infty$$
so $S$ is not bounded. Together with the fact (which you claim to know) that if $X$ is compact then $S$ is bounded, we get that $S$ is bounded if and only if $X$ is compact.
