Let $X=V(F_1,...,F_k)\subset \mathbb{P}^n$with $F_i\in k[X_0,...,X_n]$ an projective algebraic set. Let $C(X)\in \mathbb{A}^{n+1}$ the affine cone over $X$, that is
$C(X)=\theta^{-1}(X)\cup \{(0,...,0)\}$
where $\theta:\mathbb{A}^{n+1}\setminus \{(0,...,0)\}\rightarrow \mathbb{P}^n$ is defined by $P\to \bar{P}$
I want to show that every irreducible component of the cone contains the vertex $(0,...,0)$. Any suggestion is appreciated.
This corresponds to the fact that any ideal generated by homogeneous polynomials is the intersection of homogeneous ideals. That is, if you know this (or see how to prove it), your claim will follow (because if every component is defined by homogeneous ideals, clearly $(0,\ldots,0)$ lies in every component).
Here's another way. Note that $C(X)$ comes with a natural $k^\ast$-action. For any $P \in C(X)$, the point $tP$ also lies in $C(X)$ for $t \neq 0$. Since $C(X)$ is closed, it follows that $0 \in C(X)$.
Now write $C(X) = \cup_{i=1}^n X_i$ as a union of irreducible components. I claim that there is a subgroup $L^\ast \subset k^\ast$ such that $L^\ast$ act on the $X_i$. Now since $k^\ast$ act on $C(X)$, it must permute the $X_i$. But since there are only $n$ components, all the elements $c^n$ must fix $X_i$. Hence the subgroup $\{ c^n \mid c \in k ^\ast \}$ act on each $X_i$ individually.
Each of the irreducible components are closed. Consider $P \in X_0$. Then $\{ tP \mid t \in k \}$ is an irreducible closed set contained in $X$ (it is a line), that intersects $X_0$. Now the $k^\ast$-action restrict to an action on $X_0$, so the whole line must be contained in $X_0$ since $X_0$ is closed.
Remark Note that if $k^\ast$ is a divisible group (for example if $k=\mathbb C^\ast$), then we can take $L=\mathbb C$.