First of all, let me note that the terms minimum, maximum, infimum and supremum do not live in a vacuum. They are properties that a certain element can have with respect to a given order. Now forget about this and let us only focus on the order $(\Bbb R; \le)$ that you seem to be interested in. Given any subset of real numbers $S \subseteq \Bbb R$, we may ask whether or not there is some $r \in \Bbb R$ such that $r$ is bigger then all the elements in $S$. If such an $r$ exists, we call it an upper bound for $S$.
For example: Let $S = (0,1)$. Every element $s \in S$ is smaller than $5$ and hence $5$ is an upper bound for $S$. However, there are smaller upper bounds. $4$ is an upper bound for $S$ as well and so is $\pi$ and $\sqrt{2}$. We may hence ask if there is a least upper bound for $S$ and in fact, we have that $1$ is an upper bound for $S$ and no $r < 1$ is an upper bound for $S$. Hence $1$ is the least upper bound for $S$, which we also call the supremum of $S$.
We designed the real numbers in such a way that once a given subset of reals $S \subseteq \Bbb R$ has any upper bound, it will always have a (unique!) least upper bound, i.e. a supremum. (A supremum is, by definition, the same as a least upper bound.) Let me stress that this is by construction of the reals, it is not true for other orders. For example, in $(\Bbb Q; \le)$, the set $\{ q \in \Bbb Q \mid q^2 < 2\}$ clearly has an upper bound, but it doesn't have a supremum.
Let us return to $(\Bbb R; \le)$. Note that $\{ x \in \Bbb R \mid 0 < x \}$ does not have an upper bound and thus does not have a supremum. Recall that any subset $S \subseteq \Bbb R$ has a supremum if and only if it has an upper bound.
So what about maxima? A maximum of a set of reals $S \subseteq \Bbb R$ is a least upper bound of $S$ that is also an element of $S$. Since suprema are unique (provided they exist), we thus have that a maxima of $S$ is also the supremum of $S$ and that $S$ has at most one maximum.
Let's have a second look at $S = (0,1)$. We already know that $S$ has $1$ as it's suprema, but $1 \not \in (0,1)$. This immediately implies that $S$ does not have a maximum. (By the argument above, if $S$ had a maximum, it would be equal to its supremum. But the supremum of $S$ is not an element of $S$ and hence not a maximum.)
I'll leave the case for infima of $S \subseteq \Bbb R$ (which are greatest lower bounds of $S$) and minima of $S$ (which are greatest lower bounds of $S$ that are also elements of $S$) to you.
You want to show that for all $x<1$, $x$ is not an upper bound of $A$, which means there exists some $a\in A$ such that $a>x$. Finding this $a$ will have to depend on $x$.
Similarly, you want to show that for all $x>0$, $x$ is not a lower bound of $A$, meaning there exists some $a\in A$ such that $a<x$. Finding this $a$ will be easy.
Best Answer
Every element of the set $C:=\{c\in\mathbb R\mid\forall x\in A[x>c]\}$ is by definition a lower bound of $A$.
Also by definition $a$ is the greatest lower bound (=infinum) of $A$ so we must have $c\leq a$ for every $c\in C$.
That makes $a$ an upper bound of $C$.
It now remains to prove that $a$ is the least upper bound (=supremum) of $C$.
For every $x<a$ some $y$ exists with $x<y<a$.
From $y<a$ it follows easily that $y\in C$ and then from $x<y$ it follows that $x$ is not an upperbound of $C$.
Proved is now that $a$ is the least upper bound of $C$.
Edit:
Conversely let it be that $a=\sup C$.
If $x\in A$ then clearly $x$ is an upper bound of $C$ and we conclude that $a\leq x$.
This tells us that $a$ is a lower bound of $A$.
It remains to be shown that $a$ ist the greatest lower bound of $A$ i.e. that every $y>a$ is not a lower bound of $A$.
If $y>a$ is a lower bound of $A$ then for every $x<y$ we find that $x\in C$.
So taking $y>x>a$ then leads to $x\in C$ and this contradicts $a=\sup C$.
We conclude that condition $y>a$ makes it impossible for $y$ to be a lower bound of $A$.
Proved is now that $a=\inf A$.