Proving the snake lemma without a diagram chase Suppose we have two short exact sequences in an abelian category
$$0 \to A \mathrel{\overset{f}{\to}} B \mathrel{\overset{g}{\to}} C \to 0 $$
$$0 \to A' \mathrel{\overset{f'}{\to}} B' \mathrel{\overset{g'}{\to}} C' \to 0 $$
and morphisms $a : A \to A', b : B \to B', c : C \to C'$ making the obvious diagram commute. The snake lemma states that there is then an exact sequence
$$0 \to \ker a \to \ker b \to \ker c \to \operatorname{coker} a \to \operatorname{coker} b \to \operatorname{coker} c \to 0$$
where the morphisms between the kernels are induced by $f$ and $g$ while the maps between the cokernels are induced by $f'$ and $g'$.
It is not hard to show that the morphisms induced by $f, g, f', g'$ exist, are unique, and that the sequence is exact at $\ker a, \ker b, \operatorname{coker} b, \operatorname{coker} c$. With the use of a somewhat large diagram shown here, we can even construct the connecting morphism $d : \ker c \to \operatorname{coker} a$. However, I'm stuck showing exactness at $\ker c$ and $\operatorname{coker} a$. I thought Freyd might have had an element-free proof in his book, but it turns out he proves it by diagram chasing and invoking the Mitchell embedding theorem [pp. 98–99]. Is there a direct proof?
 A: The salamander lemma described by George Bergman and summarised by nlab & the secret blogging seminar help simplify proofs of the basic diagram chases, including the 3x3, four, snake and long exact diagrams.
However the Secret Blogging Seminar says: 

If you don’t like diagram chases, it’s likely that you still won’t like them once you know the Salamander lemma. The salamanders chase the diagrams for you, but you still have to chase the salamanders. I think the salamander proofs are easier to explain (once you know the Salamander lemma), and it’s easier to see where you use the hypotheses. For example, it is totally clear that the argument for the 3x3 lemma can prove the 20x20 lemma as well.

A: You can always "diagram chase" in any abelian category, without invoking any embedding theorem, using arguments with subobjects, as in MacLane's book.  
In any case, you can also construct the boundary map as follows:
We are given a map $b: B \to B'$.  Let $B'' \hookrightarrow B$ denote
the preimage in $B$ of $\ker c$.  (If you want to desribe this in more categorical
terms, it is the kernel of the composite $B \to C \to C'$.)
Then the map $B''\hookrightarrow B \rightarrow B'$ factors through the monomorphism $A' \hookrightarrow B'$
(using the fact that $A' =\ker(B' \to C')\, \, $).
This then induces a map on quotients $ B''/A \to A'/\operatorname{im}A$,
which is precisely the desired map $\ker c \to\operatorname{coker}a.$
Checking the various exactness claims is just a matter of using all the relevant universal properties of kernels, cokernels, quotients, etc. 
A: There is very nice construction of connecting morphism in Borceux: Handbook of categorical algebra II., ch. 1.09 & 1.10. Then he proves exactness of the sequence using pseudo-elements, a technique that makes diagram chasing in any abelian category similar to the diagram chasing in the categories of modules over a ring (without getting lost with all that universal properties).
Nothing non-trivial is required to understand that proof, and surely not Freyd-Mitchell embedding theorem (which is proved later in the book).
A: There's a completely element-free proof in Kashiwara's "Categories and Sheaves" (Section 12.1).
