Proof that Rel is a Category The Awodey book about Category Theory gives this definition for Rel:

The objects of $\text{Rel}$ are sets, and an arrow $A → B$ is a
  relation from $A$ to $B$, that is, a subset $R ⊆ A×B$. 
The equality relation $ \{ \langle a, a\rangle ∈ A×A\;|\; a ∈ A\}$ is the identity
  arrow on a set $A$. 
Composition in $\text{Rel}$ is to be given by: $$ S ◦ R = \{\langle a, c \rangle ∈ A × C \; |\; ∃b \;(  \langle a, b\rangle ∈ R \; \text{ and }\; \langle b, c\rangle ∈ S) \}$$
for $R ⊆ A × B$ and $S ⊆ B × C$.

How can I actually prove that Rel is a category? 
I'm not really into categories yet but this seem quite obvious since I know a category must have objects (that are the sets, in this case), arrows (relations), identity function (equality relation) and composition of functions (that is the association of relations).

Even though this question was posted quite some time ago I feel like this exercise is very useful for anyone getting into category theory, which is why I provided a solution.
As Eric said, to show that $\mathbf{Rel}$ is a category we need to show that the identity arrows satisfy the identity law and the composition satisfies the associativity law.
Let $f: A \to B$ be a subset $f \subseteq A \times B$, $g: B \to C$ be subset $g \subseteq B \times C$, and $h: C \to D$ be a subset $h \subseteq C \times D$.
To show that $f \circ 1_A = f$, we use the axiom of extensionality (for both are sets).
For the forward direction, let $(a, b) \in f \circ 1_A$. Then we see by definition that there exists some $p$ s.t. $(a, p) \in 1_A$ and $(p, b) \in f$. Since $(a, p) \in 1_A$ implies that $p=a$, we have that $(a, b) \in f$. For the backward direction, let $(a, b) \in f$. Moreover, since $(a, a) \in 1_A$, we have per definition that $(a, b) \in f \circ 1_A$. The argument that $1_B \circ f = f$ is similar.
For the associativity, we want to show that $(h \circ g) \circ f = h \circ (g \circ f)$. We know that
\begin{align*}
(a, d) \in (h \circ g) \circ f &\iff \exists b: (a, b) \in f, (b, d) \in (h \circ g) \\
&\iff \exists b, c: (a, b) \in f, (b, c) \in g, (c, d) \in h \\
&\iff \exists c: (a, c) \in (g \circ f), (c, d) \in h \\
&\iff (a, d) \in h \circ (g \circ f),
\end{align*}
which is what we wanted to show.
