With diagram chasing.
Assume, for instance, that $A$ and $C$ are exact and let's show that so is $B$. Call the morphisms between the complexes $\alpha : A \longrightarrow B$ and $\beta : B \longrightarrow C$.
Let $b \in B_n$ be a cycle; that is, $db = 0$. Consider $\beta_n b \in C_n$. Because $\beta$ is a morphism of complexes,
$ d\beta_n b = \beta_{n-1} db = \beta_{n-1} 0 = 0. $
So $\beta_n b$ is a cycle, but $C$ is exact, hence there exists $c \in C_{n+1}$ such that $dc = \beta_n b$. Since $\beta_{n+1} : B_{n+1} \longrightarrow C_{n+1} $ is an epimorphism, there exists b' \in B_{n+1} such that \beta_{n+1}b' = c.
Now, consider the element b-db' \in B_n. Since
\beta_n (b-db') = \beta_n b - \beta_ndb' = \beta_n b - d\beta_{n+1}b' = \beta_n b -dc = 0 \ ,
we have b-db' \in \ker \beta_n = \mathrm{im}\ \alpha_n. So, there is an element $a \in A_n$ such that \alpha_n a = b - db'.
This $a$ is a cycle:
\alpha_{n-1}da = d\alpha_n a = db - d^2b' = 0 \ ,
and, since $\alpha_{n-1}$ is a monomorphism, this implies $da = 0$.
But $A$ is exact. Hence there is some a' \in A_{n+1} such that da' = a.
Consider the element b'+\alpha_{n+1}a' \in B_{n+1}:
d(b'+\alpha_{n+1}a') = db' + d\alpha_{n+1}a' = db' + \alpha_nda' = db' +\alpha_n a = db'+(b - db') = b \ .
So $b$ is also a boundary and we are done.