The following proof that any artinian ring $R$ has $\text{dim}(R)=0$ is from Theorem 8.1 of Atiyah Macdonald:
Let $\frak{p}$ be a prime ideal of $R$, so that $S=R/\frak{p}$ is an artinian domain (this is because any infinite descending chain of ideals in $S$ could be lifted to an infinite descending chain of ideals in $R$, so because $R$ is artinian, $S$ must be artinian too). For any non-zero $x\in S$, we must have that $(x^n)=(x^{n+1})$ for some $n$ (this is because $S$ is artinian, so we can't have an infinite descending chain $S\supset (x)\supset (x^2)\supset\cdots$), hence $x^n=x^{n+1}y$ for some $y\in S$, but because $S$ is a domain, we can cancel to get $1=xy$, hence $x$ is invertible. Any non-zero element of $S$ is invertible, hence $S=R/\frak{p}$ is a field, hence $\frak{p}$ is maximal. Because any prime ideal of $R$ is maximal, we must have that $\text{dim}(R)=0$.
Theorem 8.5 of Atiyah Macdonald says that $R$ is artinian $\iff$ $R$ is noetherian and $\text{dim}(R)=0$.
So here is an example of a ring $R$ with $\text{dim}(R)=0$ but $R$ not noetherian, and hence not artinian:
$$R=k[x_1,x_2,\ldots]/(x_1,x_2,\ldots)^2\cong k[\epsilon_1,\epsilon_2,\ldots]$$
which is a field $k$ with infinitely many nilpotent elements $\epsilon_i$ added in. There's only one prime ideal of $k$, namely the zero ideal, and nilpotents won't change that, so the ideal $(\epsilon_1,\epsilon_2,\ldots)$ is the only prime ideal of $R$, but $R$ is certainly not noetherian - the chain of ideals
$$(0)\subset (\epsilon_1)\subset(\epsilon_1,\epsilon_2)\subset\cdots$$
is an infinite ascending chain.
Hint:
For any $n\in\mathbf N$, there exists a prime ideal of height $n$. Namely,
$$(X_1, X_2,\dots, X_n)\varsupsetneq(X_1, X_2,\dots, X_{n-1})\varsupsetneq\dots\varsupsetneq(X_1, X_2)\varsupsetneq (X_1)\varsupsetneq\{0\}$$
is a chain of prime ideals with length $n$.
Best Answer
It is a chain of length one. But it isn't a chain of prime ideals. And for Krull dimension we look only at chains of prime ideals.
For a (commutative) ring $R$ the ideal $\{0\}$ is prime if and only if $R$ is an integral domain. And an Artinian ring is an integral domain if and only if it is a field. So if $R$ is Artinian but not field then $\{0\}$ is not prime. On the other hand if $R$ is a field than it has no proper ideal. So the chain $\{0\}\subsetneq P$ is never a chain of prime ideals when $R$ is artinian.
Also note that fields are special Artinian rings. And indeed they all have Krull dimension $0$.