Is this a poset? Is $(S, R)$ a poset where $S$ is the set of all people in the world and $(a, b) \in R$, where $a$ and $b$ are people if $a$ is no shorter than $b$?
My attempt:
$a$ is no shorter than $a$. This is not reflexive because $a$ can be taller than $a$. Contradiction
True?
 A: Are you shorter than yourself? I kind of doubt it. Thus, $\langle\text{Don Larynx},\text{Don Larynx}\rangle\in R$. The same goes for anyone else, so $R$ is reflexive.
You may be having trouble with no taller than: $a$ is no taller than $b$ if the height of $a$ is $\le$ the height of $b$. From this it’s also not hard to check that $R$ is transitive. 
The crucial property is antisymmetry. What happens if there are two different people whose heights are equal?
A: False. How can $a$ be taller than himself? Every person is exactly of their own height, and not an atom taller. Therefore every person is not shorter than themselves. So the relation is reflexive.
It is also transitive, because if I am not shorter than you, and you are not shorter than my dad, then I am certainly not shorter than my dad.
However, this fails to be anti-symmetric. What happens if we are both of the same height?
A: Let's start with some modelling. 
Let $P$ be a set, $l:P\rightarrow\mathbb{R}$
a function and $R\subset P\times P$ the relation defined by $pRq\iff l\left(p\right)\geq l\left(q\right)$
Clearly the relation is reflexive and transitive, so it is a preorder.
The relation is anti-symmetric if and only if $f$ is injective. This
shows that $R$ is a partial order if and only if $f$ is injective.
To apply it here identify $P$ with the 'set of all people in the
world' and let $l$ send each person $p\in P$ to its length $l\left(p\right)$.
