Let $X \subset \mathbb{P}^n$ be a nonempty projective variety. Show that the dimension of the cone $C(X):=\{0\} \cup \{(x_0,...,x_n)\in \mathbb{A}^{n+1}:(x_0:...:x_n)\in X\}$ is dim$X+1$.

I know how to prove dim$C(X) \geq$ dim$X+1$: Let $\varnothing \neq X_0 \subsetneq ... \subsetneq X_n\subset X$ be a chain of irreducible closed subsets in $X$, then $\{0\} \subsetneq C(X_0) \subsetneq ... \subsetneq C(X_n) \subset C(X)$ is a chain of irreducible closed subsets in $C(X)$.

However I cannot prove the other side of the inequality, if I let $\varnothing \neq Y_0 \subsetneq ... \subsetneq Y_m\subset C(X)$ be a chain of irreducible closed subsets in $C(X)$, I do not know how to generate a chain of irreducible closed subsets in $X$. If the $Y_i$ are of form $V(S_i)$, where $S_i$ is a homogenous ideal, then I can take projectivization $\mathbb{P} (Y_i):=\{(x_0:...:x_n)\in \mathbb{P}^{n+}:0 \neq (x_0,...,x_n)\in X\}$ and obtain my conclusion easily. Is there a way to assert this? This is equivalent to the claim that every irreducible component of a cone is also a cone, which seems to have some geometric basis.

I have seen some answers on related questions on this site using the tool of fiber dimension and transcendence degree of the coordinate ring, however I do not want an answer with these advanced techniques, as the notes I am reading does not assume the reader has these knowledge; it would be very nice to see a proof using basic constructions such as cone and projectivization defined above.

Any help is appreciated!