Working out $\operatorname{Proj} k[x_0,...x_n]/(x_0^2,x_0x_1)$ Let $I$ be the homogeneous ideal given by $(x_0^2,x_0x_1)$ and describe $X= \operatorname{Proj} k[x_0,...x_n]/I$.
My goal here is to see that even though $I$ is not a radical ideal, X is still a reduced scheme.*
The points of $X$ are the prime ideals containing $I$ that do not contain the irrelevant ideal $(x_0,...,x_n)$. Or, what is more geometric, one can picture the locus in $\Bbb A_k^{n+1}$ where $I$ vanishes, throw away the origin and quotient by the action of $k$. $V(I)$ is the intersection of the hyperplane $x_0=0$ "with some fuzz" with the "cross" $x_0x_1=0$. Set theoretically, this is just the hyperplane $x_0=0$. We proceed to figure out what it is scheme-theoretically, i.e. we must find out if there is any fuzz and where it is.
Expanding any polynomial as a Taylor series anywhere at the hyperplane $x_0=0$, we see that after quotienting out by the relations $x_0^2=0=x_0x_1$, we still get all the partial derivatives in all directions that do not involve $x_0$. Among those that involve $x_0$, we lose those that have $x_0^2$ and those that have $x_0$ and $x_1$ (although we still get some "part" of these terms if we are evaluating somewhere at $(x_1-a)$, where $a\neq0$). Now this is messy and I don't know how to put all this fuzz in different directions on the picture, but clearlt the set $V(I)$ is not reduced at all in $\Bbb A_k^{n+1}$ since (some of) its fuzz is going out in the direction perpendicular to the hyperplane $x_0=0$.
Now, the crutch of the matter is that, in $\Bbb P_k^{n}$, the stalks are given by first localizing and then doing the additional step of looking at the degree $0$ part. So I should look at $((k[x_0,...,x_n]/I)_{(x_0,...)})_0$, but this is still non-reduced, right? Say I'm looking at a stalk where I am allowed to invert $f=x_3-4x_n$. Then I have the element $\frac {x_0} {f}$ in my local ring and its square is zero.
*EDIT: Let me quote Vakil's lecture notes p. 158 for those who asked why it should be reduced

Caution: The example of $I=(x_0^2,x_0x_1)$ shows that $\operatorname{Proj} k[x_0,...,x_n]/I$ can be a projective $k$-variety without $I$ being radical.

 A: I think $X$ is not generally reduced. Let's look at the case $X = \operatorname{Proj}(k[x_0,x_1,x_2]/(x_0^2,x_0x_1))$.
We examine this scheme in the affine chart $U$ defined by $x_2 \neq 0$. Since localization and taking degree zero pieces both commute with quotients, we have $X\cap U = \operatorname{Spec}(k[x,y]/(x^2,xy))$. That is, we invert $x_2$, set $x = x_0/x_2$ and $y = x_1/x_2$ to be coordinates on $U$, and obtain the local generators for the ideal, which "don't change" since they have no $x_2$ terms. (We are dehomogenizing the generators with respect to $x_2$.) The ideal of $X\cap U$ is $(x^2,xy) = (x)\cap (x,y)^2$, which implies that $X\cap U = \operatorname{Spec}(k[x,y]/(x,y)^2) \cup \operatorname{Spec}(k[x,y]/(x))$. This is the union of a (reduced) line with an embedded order two fat point, which is nonreduced. This shows exactly where the nonreducedness lies, since if localize at $x_1\neq 0$, then we get a reduced line.
If $n>2$, a similar phenonmenon occurs, except that line becomes a hyperplane, and the embedded point becomes an embedded codimension $2$ fat linear subscheme. (What happens if $n=1$?)
