Rigorous version of heuristic argument for genus-degree formula? A recent MO question about non-rigorous reasoning reminded me of something I've wondered about for some time.
The genus–degree formula says that genus $g$ of a nonsingular projective plane curve of degree $d$ is given by the formula $g=(d−1)(d−2)/2$. Here is a heuristic argument for the formula that someone once told me. Take $d$ lines in general position in the plane; collectively these form a (singular) degree-$d$ curve. There are $d\choose 2$ points of intersection. Now think in terms of complex numbers and visualize each line as a Riemannian sphere. If you start with $d$ disjoint spheres and then bring them together so that every one touches every other one (deforming when necessary) then you expect the genus of the resulting surface to be ${d\choose 2}−(d−1)=(d−1)(d−2)/2$, because after you connect them together in a line with $d−1$ connections, each subsequent connection increases the genus by one.

Is there any rigorous proof of the genus–degree formula that closely follows the above line of argumentation?

A standard proof of the  genus–degree formula proceeds by way of the adjunction formula.  This doesn't seem to me to answer my question, but perhaps I just don't understand the adjunction formula properly?
 A: Yes, this argument can be made rigorous. One needs three steps.
Step 1. Show that there is at least one smooth plane curve of degree $d$ with the expected genus. Essentially, the proof is given by your heuristic topological argument (deform the union of $d$ lines in general position).
Step 2. Show that if one slightly perturbs the coefficients of a homogeneous polynomial defining a smooth curve, the genus remain unchanged. This is basically a continuity argument.
Step 3. Show that the space $\mathbb{C}^{\rm nonsing}[x,\,y,\,z]_d$ of homogeneous polynomials of degree $d$ in three variables defining smooth curves is path-connected. This is because the complement of $\mathbb{C}^{\rm nonsing}[x,\,y,\,z]_d$ in $\mathbb{C}[x,\,y,\,z]_d$ (the so-called "discriminant locus") has real codimension $2$.
Putting these three steps together one easily obtains the desired result. For further details you can look at Chapter 4 of Kirvan's book Plane algebraic curves.  
A: Of course it seems you meant topological genus, but here is a similar argument for the arithmetic genus, that I learned from a paper of Fulton.  Maybe this is related to the adjunction argument you refer to.  And if you combine this with Francesco's answer, you get a proof that the two are equal for smooth plane curves, (or if you assume that result, the Hirzebruch-Riemann-Roch theorem for curves, then you get an answer to your question).
Lemma A:  If $X,Y$ are two curves on a smooth surface $S$, and if $X,Y$ are linearly equivalent as divisors on $S$, then $\chi(\mathcal{O}_X) = \chi(\mathcal{O}_Y)$.
Remark:  This says in some sense $\chi(\mathcal{O})$ is a deformation invariant, at least for linear deformations.
Proof:  Since the line bundles $\mathcal{O}_S(-X)$ and $\mathcal{O}_S(-Y)$ are isomorphic on $S$, the invariants $\chi(\mathcal{O}_S(-X))$ and $\chi(\mathcal{O}_S(-Y))$ are equal.  By the usual exact sheaf sequence $$0\to  \mathcal{O}_S(-X)\to \mathcal{O}_S\to \mathcal{O}_X\to 0$$ and the analogous one for $Y$, plus the additivity of $\chi$, we get that 
\begin{align}
\chi(\mathcal{O}_X) 
&= \chi(\mathcal{O}_S) - \chi(\mathcal{O}_S(-X)) \\
&= \chi(\mathcal{O}_S) - \chi(\mathcal{O}_S(-Y)) \\
&= \chi(\mathcal{O}_Y).\hspace{1cm}&\text{qed.}\end{align}
Lemma B:  Now suppose that $Y, Y'$ are curves on a smooth surface $S$, and that $Y$ and $Y'$ meet transversely at precisely $n$ points.  Then we claim $\chi(\mathcal{O}_{Y+Y'}) = \chi(\mathcal{O}_Y) + \chi(\mathcal{O}_{Y'}) - n.$
Proof:  Consider the sequence 
$$0\to \mathcal{O}_{Y+Y'}\to \mathcal{O}_Y + \mathcal{O}_{Y'}\to O_{Y\cdot Y'}\to 0,$$ induced by the map from the disjoint union of $Y,Y'$, to their union $Y+Y'$ on $S$, and where the map to $\mathcal{O}_{Y\cdot Y'}$ is the difference of the two restrictions, from $Y$ and from $Y'$, to the intersection of $Y$ and $Y'$.  The additivity of $\chi$ then implies the desired relation, i.e. 
\begin{align}
\chi(\mathcal{O}_Y) + \chi(\mathcal{O}_Y') 
&= \chi(\mathcal{O}_Y + \mathcal{O}_{Y'}) \\
&= \chi(\mathcal{O}_{Y+Y'}) + \chi(\mathcal{O}_{Y\cdot Y'}) \\
&= \chi(\mathcal{O}_{Y+Y'}) + n.
\end{align}  Thus $\chi(\mathcal{O}_{Y+Y'}) = \chi(\mathcal{O}_Y) + \chi(\mathcal{O}_{Y'}) - n.$  qed.
Now that we know how the function chi(O) behaves under linear degeneration, all we need is to find a formula that behaves this way, and it must be the formula for $chi(\mathcal{O})$.  
Corollary:  If $X$ is a smooth plane curve of degree $d$, then 
$\chi(\mathcal{O}_X) = 1 - \frac12 (d-1)(d-2).$
Proof:  Induction on $d$.  If $d = 2$, then the smooth conic $X$ moves in a linear series also containing a union $Y$ of two lines $Y_1 + Y_2$, where each line is isomorphic to $X$.  Then by lemmas A,B above, we have 
$$\chi(X) = \chi(Y_1)+\chi(Y_2) - 1 = \chi(X)+\chi(X)-1,$$ hence $\chi(X) = 1$.  This proves the case $d = 2$, and since a smooth curve of degree $d = 1$ is isomorphic to one of degree 2, we also obtain the formula for degree $d=1$.
Now assume $d \geq 3$ and that we have proved the formula for smooth curves 
of degree $<d$.  A smooth degree-$d$ curve $X$ moves in a linear series that also 
contains a curve of form $Y = Y_1+Y_2,$ where $Y_1$ is smooth of degree $d-1$, and $Y_2$ is a line meeting $Y_1$ transversely in $d-1$ distinct points.  Then lemmas A, B and induction give us that 
\begin{align}
\chi(\mathcal{O}_X) 
&= \chi(\mathcal{O}_Y) \\
&= \chi(\mathcal{O}_{Y_1})+\chi(O_{Y_2})-(d-1) \\
&= 1-\frac12 (d-2)(d-3) + 1 - (d-1) \\
&= 1-\frac12 (d-1)(d-2),\end{align} as desired.  qed.
