let $f_1(x,y),f_2(x,y),\dots{}$ be an infinitely countable number of polynomials in $\mathbb{C}[x,y]$. Denote by $C_n$ the curve in $\mathbb{C}^2$ given by $$ C_n = \bigl\{ (a,b) \in \mathbb{C}^2 \mid f_n(a,b) = 0 \bigr\}. $$
If $C$ is the intersection of all curves, that is,$$ C= \bigcap^{\infty}_{n=1}C_n$$ and the ideal $I$ is generated by $f_1(x,y),f_2(x,y),\dots{}$ is different than $\mathbb{C}[x,y]$. Prove that $C \neq \varnothing$.
What I thought: Since $\mathbb{C}$ is a field, we have that $\mathbb{C}$ is Noetherian. By the Hilberts Basis Theorem, $\mathbb{C}[x,y]$ is also Noetherian.
Thus, we have that there exists $m \in \mathbb{N}$ such that $I=(f_1,\dots{},f_m)$.
I don't know how to argue that $C \neq \varnothing$.
This is a consequence of the Hilbert Nullstellensatz.
One version of the Hilbert Nullstellensatz says that every maximal ideal of $\mathbb C[x, y]$ is of the form $(x - a, y - b)$, where $a, b \in \mathbb C$. So if $\mathfrak m$ is a maximal ideal of $\mathbb C[x, y]$, then its vanishing locus $Z(\mathfrak m)$ is non-empty: $Z(\mathfrak m)$ is the singleton set containing the point $(a, b ) \in \mathbb C^2$.
A consequence of this is that if $I$ is any proper ideal of $\mathbb C[x, y]$, then its vanishing locus $Z(I)$ is non-empty. To see this, observe that $I$ is contained in some maximal ideal $\mathfrak m$. We have that $Z(I) \supset Z(\mathfrak m)$, and $Z(\mathfrak m)$ is non-empty by our discussion above.
Now in your case, $C = Z(I)$, where $I = (f_1, f_2, f_3, \dots)$ is a proper ideal. ($I$ is a proper ideal because $I \neq \mathbb C[x, y]$ by assumption.) So $C$ is non-empty.