This is related to Mumford, Red book for schemes pg 26 Prop 1 of Sec 1.5.
$X$ is a prevariety over $k$(assuming algebraically closed) if $X$ is connected and $X$ is covered by finite number of affine varieties.
Prop.1 Every prevariety is irreducible as topological space.
Basically the proof goes as the following. Say $U\subset X$ is open. It suffices to show $U$ intersects any affine open of the given covering. Consider $U'$ open set formed by the affine open sets intersecting $U$ and $V'$ open set formed by the affine open sets not intersecting $U$. Now $U'\cup V'=X$. It is clear that $U'$ is non-empty by $U\subset U'$. It suffices to show $U'=X$ which will deduce any affine open set of the covering intersects $U$ non-trivially.
Take $y\in U'\cup V'$. Then there are affine open $W_i\ni y$ from the covering s.t. $W_1\cap U\neq\emptyset$ and $W_2\cap U=\emptyset$. Now $y\in W_1\cap U$. Hence $W_1\cap U$ dense in $W_1$ by $W_1$ affine. There is a non-trivial open set $S$ of $W_1$ in $W_1\cap U$. Now consider $y\in W_1\cap W_2$. Clearly $S\cap W_2\subset W_1\cap W_2$ and $S\cap W_2\neq\emptyset$ by $y\in S\cap W_2$ and $W_2$ affine. This contradicts $y\in U'\cap V'$. Hence $U'\cap V'=\emptyset$. This yields $U'=X$ by $X$ connected.
Suppose $X$ is reducible. Then there are 2 open sets $U,V$ of $X$ s.t. $U\cap V=\emptyset$. From above argument, one sees that $U$ intersects any affine open of the given covering non-trivially. In particular, pick any $y\in V$ and its associated affine open set from the covering say $W_y\ni y$. Then $W_y\cap U\neq\emptyset$. Now $V\cap W_y$ is dense in $W_y$ by non-emptyness and $W_y$ affine. Hence $U\cap V\neq\emptyset$.
$\textbf{Q:}$ Above proof did not use finite affine variety covering or invokes finite affine variety covering. Or am I wrong here?
Best Answer
No, the proof did not use that the covering was finite.
I'm not sure what Mumford's definition of prevariety is, but I'm guessing based on the criteria given that it's an integral scheme of finite-type over $k$.
Expanding out the criteria, we have
Now aside from quasicompactness, none of the rest of the conditions require finiteness of the open cover, but quasicompactness is equivalent to the existence of a finite affine cover, which is why the finiteness assumption is made.
Edit: (For those interested)
Also in the process of reminding myself what was true, I ran across this very useful expository paper on the precise differences between irreducible, locally irreducible, and pointwise irreducible and similar questions, which among other things generalizes the proof given in the question.