Let $\alpha:C\to D$ be a morphism of chain complexes in an abelian category, and denote by $\rm con$ the mapping cone of $\alpha$. There is an exact sequence of chain complexes: $$0\to D\to\mathrm{con}\to C[-1]\to 0.$$ I'm looking for a (categorical) proof that the connecting morphism $H_{n-1}(C)\to H_{n-1}(D)$ is $H_{n-1}(\alpha)$, for any $n$. I'm baffled by the fact that I arrived to an anticommutative square, which if it was commutative would lead to a proof. In other words, my proof doesn't work because of a minus sign: do you think it's a coincidence, and my attempt is vacuous, or am I close to a proof?
I used latex diagrams where possible, but I needed an awkward diagram which I had to draw; I tried to be as clear as I could, I hope it's not a problem. Thank you for your time, now I'll expose my "proof".
In the following, all differential morphisms of any chain complex will be denoted just $\partial$ (it shouldn't cause confusion). $Z$ and $B$ are used to denote kernels and images of differentials, as usual.
By the Snake Lemma, the lower row of the diagram below is exact. $\require{AMScd}$ $$\begin{CD} @.D_{n+1}@>>> D_{n+1}\oplus C_n@>>>C_{n}@>>>0\\ @. @VVV @VVV @VVV \\ 0@>>>D_{n}@>>> D_{n}\oplus C_{n-1}@>>>C_{n-1}@>>>0\\ @. @VVV @VVV @VVV \\ @. \frac{D_n}{\partial (D_{n+1})}@>>> \frac{D_n\oplus C_{n-1}}{\partial (D_{n+1}\oplus C_n)}@>>>\frac{C_{n-1}}{\partial (C_{n})}@>>>0 \end{CD}$$ Similarly, we recover another exact sequence. $$0\to Z_{n-1}(D)\to Z_{n-1}(\mathrm{con})\to Z_{n-2}(C)$$
Then in the following diagram the rows are exact.
$$\begin{CD} @. \frac{D_n}{\partial (D_{n+1})}@>>> \frac{D_n\oplus C_{n-1}}{\partial (D_{n+1}\oplus C_n)}@>>>\frac{C_{n-1}}{\partial (C_{n})}@>>>0\\ @. @VVV @VVV @VVV @.\\ 0@>>> Z_{n-1}(D)@>>> Z_{n-1}(\mathrm{con})@>>> Z_{n-2}(C) \end{CD}$$ The left vertical morphism is the unique one such that the composition below gives $\partial$ (the other vertical morphisms are obtained analogously). $$D_n\to\frac{D_n}{\partial (D_{n+1})}\to Z_{n-1}(D)\to D_{n-1}$$
Now we can refer to the drawing: the morphisms $a$, $b$, $c$, $d$, $e$, $f$, $g$ have just been defined.
- $h$ is the cokernel of $B_{n-1}(D)\to Z_{n-1}(D)$, as usual, but it is also the cokernel of $c$; $i$, $j$ are the pushout of $f$, $h$.
- $k$ is the cokernel of $B_{n-1}(C)\to Z_{n-1}(C)$, $l$ is the kernel of $C_{n-1}\to C_{n-2}$ and $m$ is the cokernel of $C_n\to C_{n-1}$. So $n$ is the unique morphism such that $nk=ml$, and it is also the kernel of $e$.
- $o$, $p$ are the pullback of $b$, $n$.
- $q$ is the canonical insertion in the biproduct, while $r$ is the cokernel of $D_{n+1}\oplus C_{n}\to D_n\oplus C_{n-1}$. So $m=brq$ (wrong).
- $s$ is the unique morphism such that $os=k$ and $ps=rql$ (wrong).
- $t$ is the connecting morphism, i.e. the unique morphism such that $jto=idp$.
- $u$ is the kernel of $D_{n-1}\to D_{n-2}$, $v$ is $\alpha_{n-1}$ and $w$ is the unique morphism such that $uw=vl$.
- $x$ is the canonical insertion in the biproduct, $y$ is the kernel of $D_{n-1}\oplus C_{n-2}\to D_{n-2}\oplus C_{n-3}$ and $z$ is the differential. So $xu=yf$ and $ydr=z$.
The proof was going on like this. The morphism $\alpha':=H_{n-1}(\alpha)$ is the unique one such that $\alpha'k=hw$, so the unique one such that $j\alpha' k=jhw$. To show that $t=\alpha'$ we can check that $jtk=jhw$ then. On one hand we have $jtk=jtos=idps$, and on the other $jhw=ifw$. So I tried to prove that $dps=fw$, or equivalently $ydps=yfw$. On one hand we get $ydps=ydrql=zql$, on the other $yfw=xuw=xvl$. Hence I checked if $zql=xvl$, and I saw that actually $zql=-xvl$ (since in the differential $z$ the component $C_{n-1}\to D_{n-1}$ is $-\alpha_{n-1}$).
Correction after comments. In order to obtain a chain complex morphism $\mathrm{con}\to C[-1]$, such that every component $D_n\oplus C_{n-1}\to C_{n-1}$ is the canonical projection, the differentials of $C[-1]$ must be the opposite of the differentials of $C$. Therefore $m=-brq$, and $s$ satisfies $ps=-rql$. So $ydps=-ydrql=-zql$, which is indeed equal to $xvl$.
(Correction about the image: $e$ is drawn as an epi by mistake, its not epic) 
The minus sign appears because you (probably) defined the cone to have a shift in the "wrong" way. Define it instead so that the sequence has $C$ without a shift and $D$ with one, and the Koszul sign will disappear.
That is, I presume that the differential of your cone complex is $$\partial = \begin{pmatrix} d_D & -\alpha \\ 0 & -d_C \end{pmatrix}$$ on $D\oplus C[-1]$ where $-d_C$ comes from the shift $C[-1]$. If you follow the proof of the Snake Lemma, you pick an element $z\in C$ such that $dz=0$, then lift it to $(0,c)$ which has differential $(-\alpha(c),0)$, so you indeed end up with $-H(\alpha)$. Note that this makes sense since what you get is a map $H(\alpha)[-1] : H(C[-1]) \longrightarrow H(D)[-1] \cong H(D[-1])$ and the $[-1]$ introduces the correct "Koszul sign".
Instead, consider the differential to be
$$\partial = \begin{pmatrix} -d_D & f \\ 0 & d_C \end{pmatrix}$$
on $D[1]\oplus C$. If you follow the proof of the Snake Lemma, you pick an element in $z\in C$ such that $dz= 0 $, this is covered by $(0,c)$ whose differential is $(f(c),0)$, and this gives you $H(\alpha)$, which is indeed the correct map $H(\alpha) : H(C) \longrightarrow H(D[1])[-1] = H(D)$,