TITLE: Irreducible components of topological space
QUESTION [17 upvotes]: Let $X$ be a topological space. Let $\Sigma$ be the set of irreducible components of $X$. Let $X=\cup_{i\in I} X_i=\cup_{j\in J} Y_j$, $X_i,Y_j\in \Sigma $ for some index set $I,J$. $X_i$'s are distinct from each other. $Y_j$'s are distinct from each other.

I want an example such that $X$ has two distinct expression, i.e. there exist $\{X_i|i\in I\}\neq\{Y_j|j\in J\}$, such that $X=\cup_{i\in I} X_i=\cup_{j\in J} Y_j$.

Something we knew (but not helpful for this question): $X$ must not be one of the following case:

$X$ is Noetherian, then it can be uniquely written as a union of finite distinct irreducible components. 
$X$ can be written as a union of finite distinct irreducible components, then all irreducible components of $X$ are in this expression, and expression is unique.
$X$ is a scheme, since {irreducible components} 1:1
correspond {generic points}. Thus expression is unique.


Update:  4. As stated, if $X$ is Hausdorff, then every irreducible set is a single point. (Because $E$ irr.$\Leftrightarrow$ every two nonempty opens intersect.)
Thanks.

REPLY [5 votes]: This should work. Take the square $[0,1]\times [0,1]$ with this strange topology. A subset is open if it is open in the Zariski topology (no generic points!) of every vertical segment, and in the Zariski topology of the horizontal segment $[0,1]\times\{0\}$.
Then, it's clear that the irriducible components are the vertical segments and $[0,1]\times\{0\}$, but to cover the square it's enough to use the vertical ones.