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Note that a point is a tree, and a pair of points is a forest. Thus with the given definition the circle $S^1$ is a two-tree. (In fact, any finite graph is a two-tree with the given definition.) Thus all surfaces are three-trees, and more generally all $n$-manifolds are $(n+1)$-trees.

The examples given in the original post strongly suggest that this is not the desired outcome. Perhaps the following definition is closer to what you want.

1. The zero-tree is the one-point space.
2. A connected topological space $X$ is an $(n+1)$-tree if for any pair of distinct points $x, y \in X$ there is an $n$-tree $Y \subset X - \{x, y\}$ separating $x$ from $y$.

I think that life is nicer if we also add the assumption that $n$-trees are homeomorphic images of CW-complexes. With this additional assumption I believe we can prove the following.

Lemma: Suppose that $X$ is an $n$-tree. Then $X$ is simply connected.

Edit: As Geva points out, the annulus is an easy counterexample to my claimed “Lemma”. I now do not see how to strengthen the OP hypotheses without making them very combinatorial… If you are willing to do that, then Ian’s answer seems like a very natural direction to go in.

Edit 2: My answer points out one problem and proposes a solution. However Geva Yashfe (comments below) points out a further problem. I've added a yet further example, and I conclude that much stronger combinatorial control seems to be needed.

Original: Note that a point is a tree, and a pair of points is a forest. Thus with the given definition the circle $S^1$ is a two-tree. (In fact, any finite graph is a two-tree with the given definition.) Thus all surfaces are three-trees, and more generally all $n$-manifolds are $(n+1)$-trees.

The examples given in the original post strongly suggest that this is not the desired outcome. Perhaps the following definition is closer to what you want.

1. The zero-tree is the one-point space.
2. A connected topological space $X$ is an $(n+1)$-tree if for any pair of distinct points $x, y \in X$ there is an $n$-tree $Y \subset X - \{x, y\}$ separating $x$ from $y$.

I think that life is nicer if we also add the assumption that $n$-trees are homeomorphic images of CW-complexes. With this additional assumption I believe we can prove the following.

Lemma: Suppose that $X$ is an $n$-tree. Then $X$ is simply connected.

Edit 1: As Geva points out, the annulus is an easy counterexample to my claimed “Lemma”. This is because properly embedded intervals separate points in any connected surface with boundary. Their technique generalises - if $M$ is a connected $n$-manifold with boundary then properly embedded $n$-disks separate points in $M$. As a concrete example, consider $M = S^2 \times [0,1]$.

Thus there is no way to strengthen the OP hypotheses without making them very combinatorial... If you are willing to do that, then Ian’s answer seems like a very natural direction.

Note that a point is a tree, and a pair of points is a forest. Thus with the given definition the circle $S^1$ is a two-tree. (In fact, any finite graph is a two-tree with the given definition.) Thus all surfaces are three-trees, and more generally all $n$-manifolds are $(n+1)$-trees.

The examples given in the original post strongly suggest that this is not the desired outcome. Perhaps the following definition is closer to what you want.

1. The zero-tree is the one-point space.
2. A connected topological space $X$ is an $(n+1)$-tree if for any pair of distinct points $x, y \in X$ there is an $n$-tree $Y \subset X - \{x, y\}$ separating $x$ from $y$.

I think that life is nicer if we also add the assumption that $n$-trees are homeomorphic images of CW-complexes. With this additional assumption I believe we can prove the following.

Lemma: Suppose that $X$ is an $n$-tree. Then $X$ is simply connected.

Edit: As Gera points out, the annulus is an easy counterexample to my claimed “Lemma”. I now do not see how to strengthen the OP hypotheses without making them very combinatorial… If you are willing to do that, then Ian’s answer seems like a very natural direction to go in.

Note that a point is a tree, and a pair of points is a forest. Thus with the given definition the circle $S^1$ is a two-tree. (In fact, any finite graph is a two-tree with the given definition.) Thus all surfaces are three-trees, and more generally all $n$-manifolds are $(n+1)$-trees.

The examples given in the original post strongly suggest that this is not the desired outcome. Perhaps the following definition is closer to what you want.

1. The zero-tree is the one-point space.
2. A connected topological space $X$ is an $(n+1)$-tree if for any pair of distinct points $x, y \in X$ there is an $n$-tree $Y \subset X - \{x, y\}$ separating $x$ from $y$.

I think that life is nicer if we also add the assumption that $n$-trees are homeomorphic images of CW-complexes. With this additional assumption I believe we can prove the following.

Lemma: Suppose that $X$ is an $n$-tree. Then $X$ is simply connected.

Edit: As Geva points out, the annulus is an easy counterexample to my claimed “Lemma”. I now do not see how to strengthen the OP hypotheses without making them very combinatorial… If you are willing to do that, then Ian’s answer seems like a very natural direction to go in.

Note that a point is a tree, and a pair of points is a forest. Thus with the given definition the circle $S^1$ is a two-tree. (In fact, any finite graph is a two-tree with the given definition.) Thus all surfaces are three-trees, and more generally all $n$-manifolds are $(n+1)$-trees.

The examples given in the original post strongly suggest that this is not the desired outcome. Perhaps the following definition is closer to what you want.

1. The zero-tree is the one-point space.
2. A connected topological space $X$ is an $(n+1)$-tree if for any pair of distinct points $x, y \in X$ there is an $n$-tree $Y \subset X - \{x, y\}$ separating $x$ from $y$.

I think that life is nicer if we also add the assumption that $n$-trees are homeomorphic images of CW-complexes. With this additional assumption I believe we can prove the following.

Lemma: Suppose that $X$ is an $n$-tree. Then $X$ is simply connected.

Note that a point is a tree, and a pair of points is a forest. Thus with the given definition the circle $S^1$ is a two-tree. (In fact, any finite graph is a two-tree with the given definition.) Thus all surfaces are three-trees, and more generally all $n$-manifolds are $(n+1)$-trees.

The examples given in the original post strongly suggest that this is not the desired outcome. Perhaps the following definition is closer to what you want.

1. The zero-tree is the one-point space.
2. A connected topological space $X$ is an $(n+1)$-tree if for any pair of distinct points $x, y \in X$ there is an $n$-tree $Y \subset X - \{x, y\}$ separating $x$ from $y$.

I think that life is nicer if we also add the assumption that $n$-trees are homeomorphic images of CW-complexes. With this additional assumption I believe we can prove the following.

Lemma: Suppose that $X$ is an $n$-tree. Then $X$ is simply connected.

Edit: As Gera points out, the annulus is an easy counterexample to my claimed “Lemma”. I now do not see how to strengthen the OP hypotheses without making them very combinatorial… If you are willing to do that, then Ian’s answer seems like a very natural direction to go in.