Inside the space of eventually zero families, the linear subspace spanned by $\{e_i:i\ge 1\}$ is proper and has a basis equinumerous with a basis of the whole space, so "equal dimension forces equality" fails without finite dimension

False claim: if $U$ is a linear subspace of a vector space $V$ over $F$ and some basis of $U$ is equinumerous with some basis of $V$, then $U=V$.

Let $F$ be any field, let $E\subseteq F^{\mathbb{N}}$ be the linear subspace of eventually zero families and let $e_k$ be the standard unit families (The standard unit families $e_k\in F^{\mathbb{N}}$ form a basis of the linear subspace of eventually zero families: an explicit infinite basis, built with no choice principle). Put

$B:=\{e_k:k\in \mathbb{N}\},B^{\mathrm{\prime }}:=\{e_k:k\in \mathbb{N},k\ge 1\},U:=\mathrm{span}(B^{\mathrm{\prime }})\mathrm{.}$

Then

1. $U$ is a linear subspace of $E$ and $B^{\mathrm{\prime }}$ is a basis of $U$, while $B$ is a basis of $E$;
2. $B^{\mathrm{\prime }}\approx B$ (Equinumerous sets, $A\approx B$ and $A⪯B$), both being equinumerous with $\mathbb{N}$;
3. $U\mathrm{\ne }E$: the family $e_0$ lies in $E$ and not in $U$.

So a proper linear subspace can carry a basis equinumerous with a basis of the whole space, and the equality clause of If ${\dim }_{F}V=n$ and $U$ is a linear subspace of $V$, then $U$ is finite-dimensional, ${\dim }_{F}U\le n$, and ${\dim }_{F}U=n$ if and only if $U=V$ — which is stated only for a finite-dimensional ambient space — really does need its hypothesis.

No dimension is assigned to either space. $E$ and $U$ are both infinite-dimensional (Finite-dimensional vector space, and its dimension ${\dim }_{F}V$; infinite-dimensional means having no finite basis assigns no number to such a space): for $E$ this is claim 4 of The standard unit families $e_k\in F^{\mathbb{N}}$ form a basis of the linear subspace of eventually zero families: an explicit infinite basis, built with no choice principle, and for $U$ it follows from claims 1 and 2 below, since the basis $B^{\mathrm{\prime }}$ of $U$ is linearly independent and equinumerous with $\mathbb{N}$, so $U$ can have no finite spanning set and hence no finite basis, by claim 2 of If $V$ has a spanning set with $n$ elements, then every linearly independent subset of $V$ is finite with at most $n$ elements; in particular $V$ has no linearly independent subset equinumerous with $\mathbb{N}$. And claim 2 compares two specific bases through an explicit bijection, not two cardinal numbers.

• The finite case is a theorem, and this shows why it is one. If ${\dim }_{F}V=n$ and $U$ is a linear subspace of $V$, then $U$ is finite-dimensional, ${\dim }_{F}U\le n$, and ${\dim }_{F}U=n$ if and only if $U=V$ proves that in a finite-dimensional ambient space equality of dimensions forces equality of the spaces; its proof enlarges a basis of the subspace by a vector outside it and contradicts the bound on independent sets. Here the same enlargement is possible — $B^{\mathrm{\prime }}\cup \{e_0\}=B$ is independent — and contradicts nothing, because there is no finite bound to violate.

• No cardinal arithmetic is used or implied. The comparison in claim 2 is a named bijection, $k\mapsto e_{\sigma (k)}$, between two specific sets. This item does not assign a dimension to $E$ or to $U$, and it says nothing about whether any two bases of $E$ are equinumerous; that question needs cardinal arithmetic, which is not available at this point in the reading order.

• The subspace is spanned by "all but one" basis vector. Deleting a single element from an infinite basis leaves a set that is still equinumerous with the original, which is exactly the phenomenon The pigeonhole principle on $\mathbb{N}$ rules out for finite sets and for natural numbers.