Given that, $\mathbb{R}$ is complete.
$[0,1] \subset \mathbb{R}$ and is closed $\implies [0,1]$ is complete and clearly non-empty.
So we know by BCT: $[0,1]$ is second category.

Suppose $[0,1]$ is countable, then let $[0,1] = \cup_{n\in\mathbb{N}}\{x_n\}$
Singletons are closed and have empty interior $\implies$ nowhere dense.
So $[0,1]$ is first category by def *Contradiction.

Is this proof concrete enough?


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