Your proof is ok. The remark preceding it is not. A set of the form $(-\infty , p)$ is simply not bounded below. In general, if a set $S\subseteq \mathbb R$ is bounded below that it has infinitely many lower bounds, but only one greatest lower bound.
Note: the plural of infimum is infima.
I'm interpreting the question as asking to prove that given an ordered field $\mathbb F$, if $\mathbb F$ has the lub property, then $\mathbb F$ has the glb property. Or equivalentely that if all non-empty bounded above subsets of $\mathbb F$ have a supremum, then all non-empty and bounded below subsets of $\mathbb F$ have an infimum.
Ordered fields are irrelevant here, this can be generalized to any poset.
Let $P$ be a poset such that each of its non-empty and bounded above subsets has a supremum.
The goal is to prove that any non-empty and bounded below subset $A$ of $P$ has an infimum.
Being a universal statement, one way to prove it is to start by taking an arbitrary non-empty and bounded below subset $A$ of $P$.
Now consider the set $\downarrow A$, where $\downarrow A:=\{p\in P\colon \forall a\in A(p\leq a)\}$, that is, consider the set the lower bounds of $A$.
The set $\downarrow A$ isn't empty because $A$ is bounded below. It is also bounded above by any element of $A$.
Hence it's possible to use the hypothesis that any non-empty and bounded above subset of $P$ has a supremum particularized to $\downarrow A$.
Let $s:=\sup\left(\downarrow A\right)$.
Claim: $s=\inf(A)$.
Proof: It suffices to prove that $s$ is a lower bound of $A$ and that $\forall p\in \downarrow A(p\leq s)$. The latter part follows immediately from the fact that $s$ is an upper bound of $\downarrow A$. The first part follows from the fact that any element of $A$ is an upperbound of $\downarrow A$ and hence greater than the supremum of $\downarrow A$ which is $s$.
Since $A$ was an arbitrary bounded below and non-empty subset of $P$, it wasproved that $P$ has the glp property.
Best Answer
To show that a set has no infimum you must show that for each candidate $i $ for infimum, there is a smaller number in the set.
For your example, suppose $i $ is the infimum. Consider the number $-i + 3 $. $-i + 3 \in \Bbb {R} $ and thus $2 - (-i+3) = 2 + i - 3 = i - 1 < i $ is in the set and $i $ is not the infimum.