This is a consequence of the Noether normalization theorem, and some inductions. (It is not trivial.) First of all, there is a weak interpretation of the statement, which says all chains of primes in $A$ of maximal length have length $r$. The stronger interpretation is that any chain of primes in $A$ in which no further prime insertions are possible has length $r$. The stronger statement is true, but somewhat harder to prove. [The length of $P_0 \subsetneq \cdots \subsetneq P_r$ is declared to be $r$ here.]
The normalization theorem says there is a polynomial subring $B = k[y_1, \ldots , y_r]$ of $A$ so that $A$ is integral over $B$. Then the Going Up Theorem will tell you chains of primes in $B$ lift to chains in $A$, and chains of length $s$ in $A$ contract to chains of length $s$ in $B$. There are obviously chains of length $r$ in $B$, hence also in $A$. I'll prove by induction on $r$ that there are no chains of length greater than $r$ in any $A$ of trans deg $r$. If $r = 1$, this holds because $B = k[y]$ is a PID, and chains in $B$ and $A$ correspond. In general, the first non-zero prime in a maximal length chain in $B$ must be principal, generated by an irreducible polynomial $f$. But now it is easy that $B/(f)$ has transcendence degree $r-1$, so by induction chains going on up from $(f)$ in $B$ have length bounded by $r-1$.
This only proves the weak theorem. The stronger theorem uses the Going Down Theorem for integral extensions of an integrally closed domain. The polynomial ring $B$ is integrally closed, so Going Down applies. The point now is, if we have a chain in $A$ for which no insertions are possible, then by Going Down the smallest non-zero prime $P$ in that chain must intersect $B$ in a minimal prime $(f)$ in $B$. Now induction for the stronger statement applies to $B/(f) \subseteq A/P$.
I guess this isn't a 'self-contained' proof, so perhaps doesn't provide what you hoped for. But I don't believe there is any easier method.
Here's a proof since I think the current answers are possibly a little lacking (and the approach taken in the question itself seems slightly off):
As in the question, let $A = k[X_0,\dots,X_n]$, $X = V(I)$ and $H = V(f)$. For a closed subset $Z = V(J) \subset \mathbb{P}^n_k$, write $\overline{Z}=V(J)\subset \mathbb{A}^{n+1}_k$. First we proceed to show that $\overline{X}\cap\overline{H}(=\overline{X\cap H})$ has dimension at least $1$.
Notice that $\overline{X\cap H} = V(I+(f))$, and since $I$ and $f$ are homogeneous, we have $I+(f) \subset (X_0,\dots,X_n)$, so there is some prime $I+(f) \subset \mathfrak{p} \subset (X_0,\dots,X_n)$ minimal over $I+(f)$ (corresponding to an irreducible component of the affine cone over the intersection of $X$ and $H$). If $\mathfrak{p} \subsetneq (X_0,\dots,X_n)$, then $V(I+(f))$ has dimension at least $1$, as claimed. If not, $\mathfrak{p} = (X_0,\dots,X_n)$ and so by Krull's principal ideal theorem applied to $A/I$, any chain of primes between $I$ and $(X_0,\dots,X_n)$ has length at most $1$. But since $X$ has dimension at least $1$, there are homogeneous primes $I \subset \mathfrak{q}\subsetneq\mathfrak{q}'$ not containing $(X_0,\dots,X_n)$, giving a chain of primes from $I$ to $(X_0,\dots,X_n)$ of length $2$, a contradiction.
This dimension argument gives that there is a prime ideal inbetween $I+(f)$ and $(X_0,\dots X_n)$, and so the question is: is there a homogeneous prime ideal between them? There's a high tech way and a low tech way to proceed, but know that the high tech approach is really just hiding the low tech one behind some fancy language.
High tech: We have a map $\pi:\mathbb{A}^{n+1}_k\setminus\{0\}\rightarrow\mathbb{P}^n_k$, such that for any closed subset $Z$, $\pi^{-1}(Z) = \overline{Z}\setminus\{0\}$, so the cone over the intersection having dimension $1$ means that it contains a non-zero point, and so this maps to the intersection under $\pi$, meaning that the intersection is therefore non-empty.
Low tech: In the process of the $\rm{Proj}(S_{\bullet})$ construction, we get a bijection between the primes of $S_{(f)}$ and the homogeneous primes of $S_{f}$, where the map one way is given by taking the degree $0$ component. But in fact, given any prime ideal of $S_f$, taking the degree $0$ component gives a prime ideal of $S_{(f)}$ and so, by the bijection, there is some homogenous prime of $S_f$ with the same degree $0$ component. Thus, in our construction, the prime $\mathfrak{p}\subsetneq (X_0,\dots, X_n)$ lying over $I+(f)$ doesn't contain some $X_i$, and so corresponds to a prime in $A_{X_i}$ lying over $(I+(f))_{X_i}$. Then by the remarks above there is then a homogeneous prime of $A_{X_i}$ with the same degree $0$ component, which therefore also contains $(I+(f))_{X_i}$. This prime then corresponds to a homogeneous prime of $A$ (also not containing $X_i$) that contains $I+(f)$, as desired. Actually checking that all of the correspondences work out as I claim is somewhat of a hassle (it's not hard per say, but is a bit of a faff) but bear in mind that you would have to do all of this anyway if you wanted to prove that the projection map used in the high tech approach actually has the desired property, so I don't think it can really be avoided.
Best Answer
Because $(p(x))$ is a prime ideal of height 1 for irreducible $p(x)$. Namely, if $\mathfrak{p}$ is a nonzero prime ideal contained in $(p(x))$, we can find an irreducible polynomial $q(x)\in\mathfrak{p}$. So $p(x)\mid q(x)$ and hence $(p(x))=(q(x))$ since both of them are irreducible. Thus we also have $(p(x))\subseteq\mathfrak{p}$ and hence they are equal.