Zorn's Lemma requires that every chain in a partially ordered set $X$ has an upper bound. In this article Gowers uses Zorn's Lemma to find a maximal linearly independent (over $\mathbb{Q}$) subset of $\mathbb{R}$. He does this by looking at the set of all linearly independent subsets of $\mathbb{R}$ with the partial order $\subset$.
He checks that every chain has an upper bound by considering a collection $Y$ of nested linearly independent subsets of $\mathbb{R}$. He defines the upper bound as their union. Thus every chain has an upper bound and Zorn's lemma can show that a basis for $\mathbb{R}$ over $\mathbb{Q}$ exists.
But what if $Y$ is infinite? It's not too hard to think of a chain in this case that has no upper bound- just keep adjoining appropriately chosen irrational numbers to get another subset of $\mathbb{R}$ linearly independent over $\mathbb{Q}$. The chain of such subsets has no upper bound, right?
In general, when you prove that every chain in $X$ has an upper bound, you need to consider finite and infinite chains, right? So how come Gowers, in paragraph 24 (below), implicitly assumed that $Y$ is a finite collection? Obviously it is easy to find the upper bound of a finite chain.
All we have to do if we want to apply Zorn’s lemma is check that every chain has an upper bound. So let us imagine that we have a collection $Y$ of linearly independent subsets of $\mathbb{R}$ and that for any two of those sets one is contained in the other. What could serve as an upper bound? By definition it has to be a set that contains all the sets in $Y$, so it has to contain their union. We want it to be linearly independent, so the smaller it is, the better. So there is basically only one candidate to try: the union itself. Is the union linearly independent? Well, if $t_1,\dots,t_n$ belong to the union, then each $t_i$ belongs to some linearly independent set $L_i\in Y$. Because $Y$ is a chain, one of these sets $L_i$ contains all the others. If that is $L_j$, then the linear independence of $L_j$ implies that no non-trivial linear combination of $t_1,\dots,t_n$ can be zero, which proves that the union of the sets in $Y$ is linearly independent, just as we wanted. Therefore, by Zorn’s lemma, there is a maximal linearly independent set.
The assumption is not that $Y$ is finite. But in order to verify that $\bigcup Y$, as the increasing union of linearly independent sets, is linearly independent, we only need to verify that finite combinations with non-zero coefficients do not produce $0$.
So given finitely many vectors (or in this case, real numbers) in $\bigcup Y$, there is some $Y_1,\ldots,Y_n$ that each real number lies in one of the $Y_i$'s, but since this is a finite chain, there is some $Y_k$ such that $Y_i\subseteq Y_k$ for all $i\in\{1,\ldots,n\}$. And since $Y_k$ is linearly independent, the vectors we picked initially cannot produce $0$, so they cannot do so in $\bigcup Y$ either.
This is a very typical use of Zorn's lemma. We pick some property which is of finite character. So if it wouldn't hold, there would be a finite subchain whose maximal element will witness this failure.
But you are correct. We cannot just assume our chains are finite. Finite chains are always bounded, since they have a maximal element.