Weibel IHA Exercise 1.2.5 I have started to work through 'An introduction to homological algebra' by Weibel and spend more time than I want going in circles on exercise 1.2.5.
The exercise states the following:


Proof in an elementary way (not with spectral sequences or the Acyclic assembly lemma) that if $C$ be a bounded double complex with exact rows or columns then $Tot(C)$ is acyclic.


This seems to be a basic diagram chase exercise, which is why I chased with determination and wasted hours on this small exercise.
Could someone give me a hint/answer? 
 A: This is an old question, but since it took me a while to solve this problem, I will tell
you the idea by walking through an example.
Consider the double complex
\begin{array}{ccccccccccc}
    & & 0 & & 0 & & 0 & & 0 & & 
    \\ 
    & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & 
    \\ 
    0 & \leftarrow & C_{0,2} & \leftarrow & C_{1,2} 
    & \leftarrow & C_{2,2} & \leftarrow & C_{3,2}
    & \leftarrow & 0
    \\ 
    & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & 
    \\ 
    0 & \leftarrow & C_{0,1} & \leftarrow & C_{1,1} 
    & \leftarrow & C_{2,1} & \leftarrow & C_{3,1}
    & \leftarrow & 0
    \\ 
    & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & 
    \\ 
    0 & \leftarrow & C_{0,0} & \leftarrow & C_{1,0} 
    & \leftarrow & C_{2,0} & \leftarrow & C_{3,0}
    & \leftarrow & 0
    \\
    & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & 
    \\ 
    & & 0 & & 0 & & 0 & & 0 & & 
\end{array}
with exact rows.
Let us look at the total complex $T(C)$, and say we concentrate on the diagonal for $n=3$,
that is, we look at
\begin{align}
    T(C)_3 = C_{1,2} \oplus C_{2,1} \oplus C_{3,0},
\end{align}
and say we have some element $x+y+z\in T(C)_3$, with $x\in C_{1,2}$ etc.
We want to show: if $d(x+y+z) = 0$ then there is some $\xi \in T(C)_4$ such that $d\xi =
x+y+z$.
What is $d(x+y+z)$? 
It is
\begin{align}
    d(x+y+z) = d^h x + d^v x + d^h y + d^v y + d^h z + d^v z.
\end{align}
Note that then $d(x+y+z)=0$ is equivalent to the four conditions
\begin{align}
    0 &= d^h x \\
    0 &= d^v x + d^h y \\
    0 &= d^v y + d^h z \\
    0 &= d^v z,
\end{align}
and the first immediately implies there is some $a\in C_{2,2}$ such that
$$
    d^h a = x.
$$
Then, we plug it into the second condition, to get
$$
    0 = d^v x + d^h y = d^v d^h a + d^h y = d^h(y - d^v a)
$$
so that $y - d^v a = d^h b$ for some $b\in C_{3,1}$.
As you may have guessed, the third condition tells us
$$
    0 = d^v y + d^h z = d^h(z - d^v b)
$$
and this would continue on and on, but in this example, the map $d^h: C_{3,0} \to C_{2,0}$
is actually injective, so $z = d^v b$.
In our example, the fourth condition doesn't even matter.
Now, $\xi = a + b$ satisfies
$$
    d\xi = d^h a + (d^v a + d^h b) + d^v b = x + y + z, 
$$
neato!
______________________________________________________-
So, the general strategy is: 
Go to the top vertical component in the diagonal.
This exists since $C$ is bounded.
The condition of an element of $T(C)_n$ being in the kernel of $d$ in particular implies
that the bit of that element that lives in the top of the diagonal is in the kernel of
$d^h$, and $d^h$ is the differential of an exact sequence.
So you can pull it back, and then...
And eventually, this chain terminates: $C$ is bounded.
And you're done.
