Consider the $C^\star$ algebra of all complex-valued continuous functions on the 2-torus $\mathbb{T}^2$ with the sup norm. Let us consider the set $\{z^m w^n\}_{m,n\in \mathbb{Z}}$, whose span is dense in $C(\mathbb{T}^2)$. We define the projection $\phi: C(\mathbb{T}^2) \to C(\mathbb{T}^2)$, thus:
$z^m w^n \mapsto z^m w^n$ if $mn\geq 0$
$z^m w^n \mapsto 0$ if $mn < 0$
and extend it linearly on the closure.
The question is, is this map contractive?
I really don't know what would be a way to attack this problem directly. I have tried a few indirect approaches using positivity, without much success: since the map is unital, it is contractive iff it is positive. Positive elements of $C(\mathbb{T}^2)$ are not very easy to characterize, due to the absence of a Fejer-Riesz theorem in 2 variables, but still, one may write a strictly positive trigonometric polynomial in $(z,w)$ as a finite sum of absolute squares of polynomials in $(z,w)$. Annihilating the terms with opposite signs for the powers of $z$ and $w$, it is not direct to see if it is positive...
As for my background: I am familiar with basic complex analysis, the basics of $C^\star$ algebras and the theory of positive maps.
Your map is not continuous, much less contractive (so there is no way to extend it by continuity beyond the trigonometric polynomials) for the following reason. Consider the embedding of $C(\mathbb T)$ into $C(\mathbb T^2)$ in such a way that $z^m$ goes to $z^mw$.
If your map is bounded, it's restriction to the above copy of $C(\mathbb T)$ would send $z^m$ to itself if $m\geq 0$, and to zero otherwise, and it is well known that this map doesn't exist on $C(\mathbb T)$.