In Munkres' Elements of Algebraic Topology Chapter 18, he aims to show that given a continuous map $h:|K|\to |L|$, there is a well-defined map $h_*:H_p(K)\to H_p(L)$, given by $f_*\circ(g_*)^{-1}$, where $f:K'\to L$ is a simplicial approximation of $h$, and $g:K'\to K$ is a simplicial approximation to the identity. But to show this is independent of the subdivision $K'$, he claims 
So I think he's saying that given two subdivisions $K'$ and $K''$ of a simpicial complex $K$, there's a common subdivision $K'''$. I see how it is enough to show for $K$ an n-simplex. But I haven't been able to show this fact. Would someone know a proof?
No, the existence of simplical approximations $k_i$ does not mean that there is a common subdivision. In fact, this is false.
It was a conjecture (the so-called Hauptvermutung, which is a German word meaning main conjecture) that any two triangulations of a polyhedron have a common subdivision. This was disproved by John Milnor in 1961. See my answer to Simplicial homology and homeomorphisms.
Update:
The OP comments that I misunderstood the question. He does not consider two arbitrary triangulations of a space $X$ and asks for a common subdivision of the triangulating complexes. He considers a simplicial complex $K$ and asks whether any two subdivisions $K', K''$ of it have a common subdivision $K'''$. I do not know the answer (although I would not be surprised it is true), but this is irrelevant for Munkres' claim. He can refer to the proof of the general simplicial approximation theorem (Theorem 16.5).
In fact, consider the identity map on $X = \lvert K \rvert = \lvert K' \rvert = \lvert K'' \rvert$. The space $X$ has two open covers $\mathcal A'$ by the open sets $St(w',K')$ and $\mathcal A''$ by the open sets $St(w'',K'')$. Choose a common refinement $\mathcal A$ of $\mathcal A'$ and $\mathcal A''$. Now choose a subdivision $K'''$ of $K$ whose closed stars refine $\mathcal A$. Then these closed stars refine both $\mathcal A'$ and $\mathcal A''$ and we get simplicial approximations $k_1 : K''' \to K'$ and $k_2 : K''' \to K''$ of $id : \lvert K \rvert \to \lvert K' \rvert = \lvert K'' \rvert$.