One natural attempt to compactify Teichmuller space is by the visual sphere of the Teichuller metric. However, Anna Lenzhen showed that there are Teichmuller geodesics which do not limit to $PMF$ (in fact, I think it was known before by Kerckhoff that the visual compactification is not Thurston's compactification).
However, it was shown by Cormac Walsh that if one takes Thurston's Lipschitz (asymmetric) metric on Teichmuller space, and take the horofunction compactification of this metric, one gets Thurston's compactification of Teichmuller space. In fact, he shows in Corollary 1.1 that every geodesic in the Lipschitz metric converges in the forward direction to a point in Thurston's boundary. I think this gives a new proof that Thurston's compactification gives a ball.
As Misha points out, it's not clear that the horofunction compactification is a ball.
Another approach was given by Mike Wolf, who gave a compactification in terms of harmonic maps, in The Teichmüller theory of harmonic maps, and showed that this is equivalent to Thurston's compactification (Theorem 4.1 of the paper). Wolf shows that given a Riemann surface $\sigma \in \mathcal{T}_g$, there is a unique harmonic map to any other Riemann surface $\rho \in \mathcal{T}_g$ which has an associated quadratic differential $\Phi(\sigma,\rho) dz^2 \in QD(\sigma)$ ($QD(\sigma)$ is naturally a linear space homeomorphic to $\mathbb{R}^{6g-6}$). Wolf shows that this is a continuous bijection between $\mathcal{T}_g$ and $QD(\sigma)$, and shows that the compactification of $QD(\sigma)$ by rays is homeomorphic to Thurston's compactification $\overline{\mathcal{T}_g}$ in Theorem 4.1. I skimmed through the proof, and as far as I can tell the proof of the homeomorphism does not appeal to the fact that Thurston's compactification is a ball, so I think this might give another proof that it is a ball.
1. Introduction
2. Easy example
3. Normality, Duality
4. Normal example
Introduction
The pseudomanifold and homology manifold conditions are both local conditions, while Poincaré duality is a global condition. It is possible for a pseudomanifold to fail the homology manifold conditions in several places, but so that the local deviations globally cancel, so it retains duality.
Simplicial complexes are "locally cone-like": every point $x$ has a neighborhood that is the cone on another space $L$, the link, with $x$ the cone point. Then $$H_*(X,X-\{x\})=H_*(CL,L\times I)=\tilde H_*(SL)=\tilde H_{*-1}(L)$$ So the homology manifold condition is that the links have the homology of spheres.
Easy example
The interval admits an involution reversing the endpoints. The quotient by this involution is again contractible. Think of this quotient operation as gluing one half of the interval to the other half. Embed the interval in an $n$-manifold. Form a quotient of the manifold by gluing the one half of the interval to the other half. If $n\geqslant 3$, this leaves the cells in degree $n$ and $n-1$ unchanged, so does not affect the pseudomanifold condition. Since we have replaced one contractible subspace by another, the homotopy type of the total space is unchanged and satisfies duality (with the pseudomanifold fundamental class, etc). But the the links are no longer spheres at any point along the interval. At the image of the ends of the interval, the link is two spheres glued in one point. At general points along the interval, it is two spheres glued in two points. At the image of the midpoint, the fixed point of the involution, the link is a single sphere with two of its points glued to each other.
Normality, Duality
That example involved changing a manifold in low dimensions, which doesn't violate the high dimensional pseudomanifold condition. To rule out such examples, one has the concept of a normal pseudomanifold (cf normal variety). The normalization of a pseudomanifold is produced by taking a disjoint union of $n$-simplices parameterized by those of the original space and gluing them along their $n-1$-faces, according to how the original was glued. Thus the normalization of the above example is the unmodified manifold. A normal pseudomanifold is one isomorphic to its normalization.
It may be useful to use the Verdier dualizing sheaf. The cohomology with coefficients in the dualizing sheaf matches the homology: $H^*(X; D)\cong H_*(X)$. An oriented pseudomanifold yields a map of sheaves $\mathbb Z\to D$. We seek Poincaré duality, that is, $H^*(X; \mathbb Z)\cong H_*(X)$. So we seek spaces where the map of sheaves induces an isomorphism on cohomology $H^*(X; \mathbb Z)\cong H^*(X; D)$, even though it is not an isomorphism of sheaves. That is, we seek the cone to be a nontrivial sheaf with no cohomology in any dimension. There are such sheaves, such as a local system supported on a circle with appropriate monodromy. That motivates the construction. It also gives immediate proofs of the claims, but it should not be hard to check them without using sheaves.
Normal example
Take an $n$-manifold manifold $M$ and an automorphism $\phi$ so that the action of on the homology in intermediate degrees is sufficiently mixing so that if we consider it an action of the group $\mathbb Z$ and take group homology with those coefficients, the homology is trivial. For example, $M=T^2$ and $\phi=\left(\begin{matrix}2&1\\1&1\\ \end{matrix}\right)$. Then form the mapping torus ($T_\phi=M\times I/(x,0)\sim(\phi(x),1)$). This is an $M$-bundle over the circle, homology computed by a spectral sequence $H_*(S^1;H_*(M))\Rightarrow H_*(T_\phi)$. By assumption on $\phi$, the spectral sequence degenerates and $H_*(T_\phi)=H_*(S^1\times S^n)$. Then cone off each copy of $M$, forming a circle of cone points. To put it another way: form the mapping cylinder of the map to the circle $T_\phi\to S^1$. Yet another way: the mapping torus of the self-map of the cone $\tilde\phi\colon CM\to CM$. This space is a normal pseudomanifold (with boundary) because those properties are preserved by cones and products. Its set of singular points is a circle, and the dualizing sheaf twists about it with the prescribed monodromy, so it contributes nothing to the sheaf cohomology. If you prefer a closed example, double the space along the boundary (or equivalently take the double mapping cylinder of the map to the circle; or the mapping torus of the automorphism of the suspension). This gives a pseudomanifold that is not a homology manifold, but which satisfies Poincaré duality. Just as the mapping torus had the homology of a manifold $S^1\times S^n$, this space has the homology of $S^1\times S^{n+1}$. Actually, that is only correct if we restrict to trivial coefficients. If we define Poincaré duality to be for all local systems, then this space fails, for some unwind the twist around the singular circles. However, we can eliminate them by killing the fundamental group by surgery. That is, cut out a neighborhood of a circle, $S^1\times D^{n+1}$, leaving a boundary $S^1\times S^n$ and fill it in with $D^2\times S^n$. Now local systems are trivial, so it satisfies full Poincaré duality.
Best Answer
The homeomorphism type of the space left after deleting the interior of the ball can depend on whether the ball intersects the singular locus (necessarily in the boundary of the ball) or not. If your ball intersects the singular locus, then once you remove the ball interior such points need not have neighborhoods homeomorphic to Euclidean half space, which will be the case if you remove the interior of a simplex in the interior of the manifold part of the pseudomanifold. On the other hand, if you restrict the balls to lie entirely disjoint from the singular locus, then your problem occurs entirely within the manifold obtained by removing the singular locus, in which case you can apply the classical results.
As a concrete example, suppose you triangulate the torus and then take your pseudomanifold to be the suspension of that torus. Let your ball be one of the cones on one of the 2-simplices of the torus. When you remove the interior of the ball, the cone point now has a neighborhood homeomorphic to the cone on a torus with an open disk removed. A local homology argument can then be used to show that the cone point does not have a neighborhood homeomorphic to Euclidean half-space.