I was looking at
Izzo, Alexander J., A functional analysis proof of the existence of Haar measure on locally compact Abelian groups, Proc. Am. Math. Soc. 115, No. 2, 581-583 (1992). ZBL0777.28006.
which proves existence of the Haar-measure for locally compact abelian groups using the Markov-Kakutani theorem.
What I find strange is that the Haar measure is constructed as an element of the dual of $C_c(X)$. But for noncompact $X$ (such as $X$ being the real numbers $\Bbb R$) this must be an unbounded functional (as the Lebesgue-measure on $\Bbb R$ is not finite). It seems like the author has no problem with this, and (without mentioning it further) goes on to define a weak-* topology for this case and even uses Banach-Alaoglu.
I have not seen this being done this way before, am I misunderstanding something or can one define a weak-* topology on the algebraic dual of a TVS without any problems?
Probably the topology on $C^o_c(\mathbb R)$ (or for other non-compact topological group in place of $\mathbb R$) is not what you anticipated. It is not sup norm, for example. It is an ascending union, categorically a "strict colimit" (=strict inductive limit=...) of spaces of the form $C^o(K)$ as $K$ ranges over compact subsets of $\mathbb R$. (These are Banach spaces.)
In particular, from the characterizing mapping property of "colimit", a linear map or functional from $C^o_c(\mathbb R)$ is continuous if and only if its restriction to each $C^o(K)$ is continuous.
So the dual of $C^o_c(\mathbb R)$ does not include "unbounded" functionals in a true sense, because even on such spaces "continuous" is still equivalent to "bounded"... but "bounded" has a more complicated sense.