• The universal covering space $\widetilde{X}$ equipped with the length metric induced by the covering projection is a length space (3.25).
• The fundamental group $\pi_{1}(X)$ acts on $\widetilde{X}$ properly and cocompactly by isometries (8.3 (2)), see also (8.5).
• If a length space $\tilde{X}$ \widetilde{X}$admits a proper and cocompact action by isometries then it is locally compact (8.4 (1)) and hence proper and geodesic (3.7). • A group is finitely presented if and only if it acts properly and cocompactly by isometries on a simply connected geodesic space (8.11). All this taken together yields that$\widetilde{X}$is a simply connected geodesic metric space and$\pi_{1}{(X)}$acts properly and cocompactly by isometries, hence$\pi_{1}{(X)}$must be finitely presented. 1 I hope I have not goofed, but I think the answer to your modified question is yes: The fundamental group of a semi-locally simply connected, compact and geodesic space is finitely presented. Here are the ingredients - all numbers in parentheses refer to Bridson-Haefliger, Metric spaces of non-positive curvature, Springer Grundlehren, 1999, Part I. • The universal covering space$\widetilde{X}$equipped with the length metric induced by the covering projection is a length space (3.25). • The fundamental group$\pi_{1}(X)$acts on$\widetilde{X}$properly and cocompactly by isometries (8.3 (2)), see also (8.5). • If a length space$\tilde{X}$admits a proper and cocompact action by isometries then it is locally compact (8.4 (1)) and hence proper and geodesic (3.7). • A group is finitely presented if and only if acts properly and cocompactly by isometries on a simply connected geodesic space (8.11). All this taken together yields that$\widetilde{X}$is a simply connected geodesic metric space and$\pi_{1}{(X)}$acts properly and cocompactly by isometries, hence$\pi_{1}{(X)}\$ must be finitely presented.