Show that $\mathbf{Ab}$ (the category of Abelian Groups) is not equivalent to its opposite category.
Here's yet another way to show it. In $Ab$, any coproduct of injective objects is injective. Concretely, the injective objects of $Ab$ are the divisible abelian groups, and any direct sum of divisible abelian groups is still divisible.
The dual property is that any product of projective objects is projective. This is false in $Ab$; for instance, a product $\mathbb{Z}^\mathbb{N}$ of infinitely many copies of $\mathbb{Z}$ is not projective (see Why isn't an infinite direct product of copies of $\Bbb Z$ a free module?).
And another, related to Hurkyl's answer. Every finitely generated abelian group $A$ is compact, meaning that the Hom-functor $\operatorname{Hom}(A,-)$ preserves filtered colimits, However, no nontrivial abelian group satisfies the dual version of compactness (that $\operatorname{Hom}(-,A)$ turns filtered limits into filtered colimits); see Can a nonzero module be cocompact?. So, for instance, $Ab^{op}$ has the property "every compact object is terminal" but $Ab$ does not.
With some knowledge of homological algebra....
One of the important properties of Ab is that direct colimits commute with finite limits. Inverse limits, however, do not generally commute with finite colimits.
Since these properties are dual, the opposite is true of $\mathbf{Ab}^{\mathrm{op}}$; in this category inverse limits commute with finite colimits, but direct colimits do not generally commute with finite limits.
The identity functor on Ab is representable by $\mathbb{Z}$ but is not corepresentable. Indeed, if $X$ were a corepresenting object, $X \cong $ Hom$(\mathbb{Z},X) \cong \mathbb{Z}$, but certainly $\mathbb{Z}$ isn't the corepresenting object. Thus Ab isn't the same as its dual.
Edit: Since I was asked for a simple proof not using representability, here is another:
A contravariant equivalence $F\colon $ Ab $\to $ Ab sends simple objects to simple objects since Ab is abelian. Thus the simple objects ($\mathbb{Z}/p\mathbb{Z}$) are permuted in such an equivalence. Since End$(\mathbb{Z}/p\mathbb{Z}) \cong \mathbb{Z}/p\mathbb{Z}$, these objects are actually fixed.
Now, the exact sequence $0 \to \mathbb{Z} \xrightarrow{p} \mathbb{Z} \to \mathbb{Z}/p\mathbb{Z} \to 0$ gets sent to $0 \to \mathbb{Z}/p\mathbb{Z} \to F(\mathbb{Z}) \xrightarrow{p} F(\mathbb{Z}) \to 0$, so the subgroup of $F(\mathbb{Z})$ killed by $p$ is $\mathbb{Z}/p\mathbb{Z}$, and $F(\mathbb{Z})$ is moreover divisible since multiplication by $p$ is surjective for each $p$. It follows that $F(\mathbb{Z})$ is $\mathbb{Q}/\mathbb{Z} \oplus \mathbb{Q}^I$, but then End$(\mathbb{Q}/\mathbb{Z}\oplus \mathbb{Q}^I) \neq$ End$(\mathbb{Z})^{op} = \mathbb{Z}$, a contradiction.