Open Unit Interval is Proper Subset of Closed Unit Interval

Theorem
The open unit interval:
 * $I_o = \left({0 \,.\,.\, 1}\right)$

is a proper subset of the closed unit interval:
 * $I_c = \left[{0 \,.\,.\, 1}\right]$

Proof
Let $x \in I_o$.

Then by definition:
 * $0 < x < 1$

and so:
 * $0 \le x \le 1$

and so:
 * $x \in I_c$.

Thus:
 * $I_o \subseteq I_c$

Consider:
 * $0 \in I_c$

by definition of closed interval.

But it is not the case that $0 < 0$.

So $0 \notin I_o$ and so $I_c \not \subseteq I_o$.

Hence the result by definition of proper subset.