Try to write down an open interval that is infinitely small in length. You'll find that you can't (unless you count the empty set, and you can't cover a closed interval in empty sets!).
Note that if you have an interval which is not closed, like $[0,1)$, then you can use a cover like $U_n = [0,\frac{n-1}{n})$, each set covering closer and closer to the right endpoint and hence eventually covering any point to the left of it. This cover has no finite subcover, since any finite collection of such sets includes a biggest one, and that biggest one misses a tiny sliver of the original interval.
What happens when you try to do the same thing with $[0,1]$? Well, you can get closer and closer and closer to $1$, but at some point you've got to hit it, so you've got to have an open interval like $(1-\epsilon,1+\delta)$ in your cover, but then you've just made infinitely many (indeed, all but finitely many) sets of your cover redundant, since you now only need to cover up to $1-\epsilon$, and regardless of how small $\epsilon$ is, there's some $1/n$ less than it, so there's an interval $[1,\frac{n-1}{n})$ that does the job.
Actually, this realisation that if you have infinitely many open sets creeping up on something, the open set actually containing that something is going to let you throw most of them away, is surprisingly close to the formal proof of your result, known as (more-or-less) the Heine-Borel theorem: just for fun, here's a proof in iambic pentameter.