In the lecture notes I have, $C_0(X)$ is defined as follows:
Let $X$ be a locally compact, Hausdorff topological space. Then $C_0(X)$ is the set of all continuous complex-valued functions on $X$ that vanish at infinity, which means that $$\lbrace x\in X:|f(x)|\geq\epsilon\rbrace$$ is compact for all $\epsilon>0$.
Here goes: Let $f,g\in C_0(X)$. We know that $f+g$ and $fg$ are continuous, and need to show that $\lbrace x\in X:|f(x)+g(x)|\geq\epsilon\rbrace$ and $\lbrace x\in X:|f(x)||g(x)|\geq\epsilon\rbrace$ are compact for all $\epsilon>0$.
I guess these sets must be expressed as some combination of $\lbrace x\in X:|f(x)|\geq\epsilon\rbrace$, $\lbrace x\in X:|g(x)|\geq\epsilon\rbrace$ so that the compactness carries over. I'm a little rusty on functional analysis so there's probably some crucial result that I've forgotten. Does $X$ being locally compact and Hausdorff need to be used?
You don't need locally compact or Hausdorff. Let $f, g \in C_0(X)$. Given any $\epsilon > 0$, we want to show that $C = \{ x \in X: |f(x) + g(x)| \geq \epsilon\}$ is compact. Let $$A = \{ x \in X : |f(x)| \geq \epsilon/3 \}$$ $$B = \{x \in X : |g(x)| \geq \epsilon/3\}.$$ These are both compact, and hence so is $A \cup B$.
The triangle inequality shows that $C$ is contained in $A \cup B$, making it compact as well. This establishes that $f+g \in C_0(X)$. A similar argument shows that $fg \in C_0(X)$.