I accept this statement as true. That's not my difficulty.
Shafarevich claims in Volume 1 of "Basic Algebraic Geometry" that it is enough to consider the case of a hypersurface in $ \mathbb{A}^{n}_{\mathbb{C}}. $ Why is this the case?
Does it have something to do with the fact that $ \mathbb{P}^{n}_{\mathbb{C}} = \bigsqcup_{i=0}^{n} \mathbb{A}^{i}_{\mathbb{C}}$ ?
Let $X$ be a hypersurface in $\mathbb{P}^n$ and let $C$ be one of its irreducible components. Then $C$ intersects one of the $n+1$ copies of $\mathbb{A}^n$ that cover $\mathbb{P}^n$; call that copy $U$. Then $U\cap C$ is a dense open subvariety of $C$ and so has the same dimension, and $U\cap C$ is an irreducible component of the hypersurface $U\cap X$ in $U\cong \mathbb{A}^n$. So, if we know the result for $\mathbb{A}^n$, we conclude that $U\cap C$ has dimension $n-1$ and thus so does $C$.