The question may require some basic knowledge in algebraic topology, which is not what I am familiar with.
$\textbf{Question}$: Do all compact connected surfaces with boundary have a finite triangulation? If the answer is affirmative, how to construct such a triangulation?
Recently I have known the result that any compact surface without boundary admits a triangulation. The compactness of the surface implies that the triangulation can be chosen to be finite. I have also learned that a compact surface with boundary can be obtained by removing a finite number of open discs from a compact surface without boundary.
I tried to construct the triangulation as follows. Let $Y$ be the compact surface with boundary that is obtained by removing disjoint open disks $\{O_i\}_{i=1}^n$ from $X$, a compact surface without boundary. That is, $Y=X \setminus (\bigcup_{i=1}^n O_i$). Write $X=\bigcup_{j=1}^mT_j$ as the union of $m$ triangles that form the triangulation. Now $Y=\bigcup_{i=1, ..., n\\j=1, ..., m}(T_j \setminus O_i)$. However, some $T_{j_0} \setminus O_{i_0}$ could be a disjoint union of infinitely many connected subspaces and each of which is homeomorphic to the closed disk, thus it seems to be difficult to construct the finite triangulation in this way.
Intuitively I believe that the answer to the question is "yes", but I wish to find a method in constructing a finite triangulation or some topics related to this.
Any suggestions would be greatly appreciated!