Why is the dimension of a ring, with finite spectrum, $\leq 1$?
It is clear that it works for the case if the ring is noetherian, due to the fact that every ideal of height $2$ is an infinite union of prime ideals of height $1$. It seems like one could use the prime avoidance lemma, but I am really unsure how to work it out.
The statement is not true if $R$ is not Noetherian. Indeed, for every $n$, there exists a ring with finite spectrum and Krull dimension $n$.
To see this, we can use valuation rings: For $n \in \Bbb N$, let $\Gamma$ be a linearly ordered group of rank $n$ (one can take for example $\Bbb Z^n$ with the lexicographical order), then there exists a valuation ring $R$ with value group $\Gamma$ (compare for example this question) and the Krull dimension of $R$ will be equal to the rank of $\Gamma$. But the prime ideals in any valuation ring are totally ordered, so for valuation rings, having finite Krull dimension is in fact equivalent to having finite spectrum.