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Edit: Your Theorem 2 is not correct.

For a counterexample, let $X = \mathbb{R}$, and let $\mathcal{T}$ be the set of all $A\subseteq \mathbb{R}$ such that $A = \mathbb{R}$ or $A\cap \mathbb{N}$ is finite. This family is stable under arbitrary intersections, and the structure $(\mathbb{R}, \mathcal{T})$ has a dimension, since $U\subseteq \mathbb{R}$ is free if and only if $U\cap \mathbb{N}$ is finite, so the union of any free set with any singleton is free.

Now let $V = \mathbb{R}\setminus \mathbb{N}$. We have $\langle \mathbb{N}\rangle = \mathbb{R}$, so $V\subseteq \langle \mathbb{N}\rangle$, and $\text{card}(\mathbb{N})< \text{card}(V)$, but $V$ is free, since $V\cap \mathbb{N} = \emptyset$.

If you were to explain your supposed proof of Theorem 2, I might be able to find where the proof goes wrong and suggest a strenthening of your axioms that makes it true.

Let me try to situation your definition in context.

Let me try to situate your definition in context.

Your definition of "has a dimension" is enough to get bases for closures of finite sets $A$ (just build up a free set by adding elements of $A$ not in the closure of what we've built so far one by one, until the closure of our basis contains $A$ and hence contains the closure of $A$). So it's enough to get exchange when the set $A$ appearing in the definition of exchange is finite. The problem comes in "taking limits" when $A$ is infinite.

This is one motivation for the definition of fintary matroid. If we assume that $\text{cl}$ is finitary (this is equivalent to assuming that the family $\mathcal{T}$ is stable under directed unions), then "has a dimension" will imply the existence of bases and exchange.

By the remark above, the finite dimensional sets in your structures with dimension will behave just like they do in finitary matroids. Thus you'll be able to prove things like the theorem in your question. But without any additional assumptions on how closures of infinite sets behave, I suspect you'll have a hard time proving anything interesting about the infinite dimensional sets.

Your definition of "has a dimension" is enough to get bases for closures of finite sets $A$ (just build up a free set by adding elements of $A$ not in the closure of what we've built so far one by one, until the closure of our basis contains $A$ and hence is equal to the closure of $A$). So it's enough to get exchange when the set $A$ appearing in the definition of exchange is finite. The problem arises when you try to "take limits": the union of an increasing sequence of free sets is not necessarily free.

This is one motivation for the definition of finitary matroid. If we assume that $\text{cl}$ is finitary (this is equivalent to assuming that the family $\mathcal{T}$ is stable under directed unions), then "has a dimension" will imply the existence of bases and exchange.

By the last remark above, the finite dimensional sets in your structures with dimension will behave just like they do in finitary matroids. Thus you'll be able to prove things like the theorem in your question. But without any additional assumptions on how closures of infinite sets behave, I suspect you'll have a hard time proving anything interesting about the infinite dimensional sets.