TITLE: In stating that the union of vector subspaces is a subspace iff they are ordered, why require $F$ finite?
QUESTION [8 upvotes]: On the bottom of page 38 of Roman's Advanced Linear Algebra is written the following (here $V$ is a vector space over the field $F$ and $\mathcal{S}(V)$ is the set of linear subspaces of $V$):
"...if $S$, $T\in \mathcal{S}(V)$ (and $F$ is infinite), then $S \cup T\in \mathcal{S}(V)$ iff $S \subseteq T$ or $T \subseteq S$."
I cannot for the life of me figure out why the finiteness of $F$ is mentioned; it seems irrelevant. The proof of $\Leftarrow$ is obvious in any case, and the proof of $\Rightarrow$ can be done by contrapositive as follows: take $s \in S \setminus T$ and $t \in T \setminus S$ and note $s+t \in T \Rightarrow s\in T$ (contradiction) and similarly for $S$, so $s+t \notin S \cup T$. 
How could this argument break down for finite $F$?

REPLY [6 votes]: As André Nicolas mentioned in the comments the argument you have also holds for finite $F$. 
What does not hold anymore is that a vector space can be a finite union of proper subspaces, e.g. $$\mathbb{F}_2^2=\langle (1,0)\rangle \cup \langle (1,1)\rangle \cup \langle (0,1)\rangle.$$ 
What does break down in your induction argument in the comments is that the union of two of these subspaces is not a subspace anymore. Thus you can not apply induction hypothesis as you have done for $(V_1\cup \dots\cup V_{n-1})$ and $V_n$ (here $V_1\cup\dots\cup V_{n-1}$ is not a subspace).