Let $C,D$ be two disjoint Zariski-closed subsets of $\mathbb{C}^n$, and let $f,g$ be polynomial functions on $C,D$ correspondingly. Then there is a polynomial function $h$ on $\mathbb{C}^n$ that agrees with both $f$ and $g$.
This is equivalent to the existence of a polynomial function $h_D$ that is $0$ on $C$ and is $1$ on $D$, since then we can just take $gh_D+fh_C$.
Let's try to prove this: We have $f \in A(C), g \in A(D)$ and $(1)=\mathbb{I}(\emptyset)=\mathbb{I}(C\cap D)= \sqrt{\mathbb{I}(C)+\mathbb{I}(D)}$, thus $\mathbb{I}(C)+\mathbb{I}(D)=(1)$ and by the chinese remainder theorem we have $A(C \cup D)=\frac{\mathbb{C}[x]}{\mathbb{I}(C)\cap \mathbb{I}(D)}\cong \frac{\mathbb{C}[x]}{ \mathbb{I}(D)} \times \frac{\mathbb{C}[x]}{\mathbb{I}(C)}=A(C) \times A(D)$.
However, I'm not convinced as to whether this actually constitutes a proof. For each $(f,g)\in A(C) \times A(D)$ we look for a function $h \in A(C \cup D)$ that agrees with $f,g$, so is that isomorphism really helpful?
Okay, you've established an isomorphism $A(C\cup D) \cong A(C)\times A(D)$, so you know how to produce a polynomial $h:C\cup D\to\mathbb{C}$ given polynomials $f:C\to\mathbb{C}$ and $g:D\to\mathbb{C}$. Presumably the thing you're wondering is: is this $h$ actually an extension of $f$ and $g$?
To see that it is, you need to actually look at how the isomorphism is constructed. You write:
$$\frac{\mathbb{C}[x]}{\mathbb{I}(C)\cap \mathbb{I}(D)}\cong \frac{\mathbb{C}[x]}{ \mathbb{I}(D)} \times \frac{\mathbb{C}[x]}{\mathbb{I}(C)}$$
An important aspect of CRT is that this is isomorphism is actually the natural map $h + \mathbb{I}(C)\cap \mathbb{I}(D) \mapsto (h+\mathbb{I}(C),h+\mathbb{I}(D))$.
So the $h$ we get from this isomorphism has to restrict to $f$ in the first component, and $g$ in the second component.