Let $(X,\leq)$ be a partially ordered set. That means that for all $x,y,z\in X$:
1) $x\leq x$ (reflexivity)
2) $x\leq y$ and $y\leq x$ imply $x=y$ (anti-symmetry)
3) $x\leq y$ and $y\leq z$ imply $x\leq z$ (transitivity)
An element $x\in X$ is a lower bound of a subset $S\subseteq X$ if $x\leq S$ for all $s\in S$. An element $x$ is the greatest element in a subset $T\subseteq X$ if $x\in T$ and $t\leq x$ for all $t\in T$. An infimum of a set $S$ is a greatest element in the set of lower bounds of $S$. We will show that there can be at most one greatest element in every set, so there can be at most one infimum for every set.
There can be at most one greatest element in a subset $T\subseteq X$.
Proof: Let $x,x'$ be both greatest elements in $T$. Then $x\in T$ and $x'\in T$ and since $x$ is a greatest element in $T$, we have $x'\leq x$. Similarly, $x\leq x'$. By anti-symmetry, $x=x'$.
Your idea is exactly right, but we often have a convention that a set with no upper bound, such as the positive real numbers, has a supremum of "$\infty$", and a set with no lower bound has an infimum of "$-\infty$". In this sense every set has a supremum and an infimum, although it may not have a minimum or a maximum.
Best Answer
The set of real numbers in $[0,1)$ isn't finite, it is bounded.
There is no real number "just below" 1. Give any real number $\alpha < 1$ and $\alpha + \frac{(1-\alpha)}{2}$ is also strictly less than 1, and strictly greater than $\alpha$.