I will answer the question submitting a proof for G. Bresciani's claim in his answer saying “it's clear the irreducible components are the vertical segments and $[0,1]\times\{0\}$.” (I don't think it's that clear; if the vertical segments and $[0,1]\times\{0\}$ were all open, then if would be immediate thanks to the bijection between irreducible components of a space and those from an open subspace, but the thing is that they are not open in this strange topology.)
Let $I$ be an infinite set. Set $X=I\times I$ and fix $i_0\in I$. Denote $V_i=\{i\}\times I\subset X$, $i\in I$ to the “vertical segments” and $H=I\times\{i_0\}\subset X$ to the “horizontal segment at $i_0$” (this terminology is motivated from the case $I=[0,1]$).
Lemma. Suppose $X$ has a topology such that:
- $V_i$ is closed and $V_i':=V_i\setminus \{(i,i_0)\}$ is open,
- the subspaces $H$ and $V_i$, $i\in I$ have the cofinite topology, and
- Either $H$ is an irreducible component of $X$ or the following property holds: if $Z\subset X$ is a closed subset such that $(Z\setminus H)\cap V_i$ is finite for all $i$, then $Z\setminus H$ is closed.
Then the irreducible components of $X$ are exactly $H$ and $V_i$, $i\in I$.
Corollary. There is a topological space with an irreducible component contained in the union of the other irreducible components.
Proof of the corollary. Suppose we defined the topology on $X$ by the rule: a subset $U\subset X$ is open if and only if, for all $i\in I$, the set $U\cap V_i$ (resp., $U\cap H$) is open in $V_i$ (resp., in $H$) with the cofinite topology. This rule defines a family of subsets that is closed under arbitrary unions and finite intersections and hence defines a topology. We verify the hypotheses of the lemma. On the one hand, points 1 and 3 are obvious. On the other hand, to see point 2, if $F\subset X$ is closed, then $F\subset V_i$ (resp., $F\cap H$) is closed in the cofinite topology of $V_i$ (resp., of $H$) by definition. Conversely, one easily verifies that $X$ is $\mathrm{T}_1$ (finite subsets of $X$ are closed). In particular, so is any finite subset contained in $V_i$ (resp., contained in $H$). Thus, we can apply the lemma to obtain $X=\bigcup_{i\in I}V_i$ is a union of irreducible components although $H$ is also another irreducible component of $X$. $\square$
Proof of the lemma. Note that $H$ is closed in $X$ (its complementary set, $\bigcup_i V_i'$, is open), and that $V_i$ and $H$ are irreducible (so is any infinite topological space bearing the cofinite topology). We will show they are irreducible components (i.e., maximal in the poset of closed irreducible subsets of $X$).
$H$ is an irreducible component (assume the ‘property’ from point 3 holds): suppose $Z\subset X$ is a closed and irreducible subset such and $H\subset Z$. Note that $Z$ does not contain $V_i$ (if it did, then $Z=V_i\cup (Z\setminus V_i')$ is a union of closed proper subsets); hence, $Z\cap V_i$ is finite. We have $Z=H\cup (Z\setminus H)$, where $Z\setminus H$ is closed (the set $(Z\setminus H)\cap V_i\subset Z\cap V_i$ is finite for all $i\in I$). Therefore, $H=Z$.
$V_i$ is an irreducible component: suppose $Z\subset X$ is a closed and irreducible subset such and $V_i\subset Z$. We know we cannot have $H\subset Z$, whence $H\cap Z$ is finite. Thus, $Z=\bigcup_{(j,i_0)\in H\cap Z} V_j\cap Z$ is a finite disjoint union of closed subsets; hence, $V_i=Z$.
Finally, we show that $H$ and $V_i$, $i\in I$, are all the irreducible components of $X$. Suppose $Z\subset X$ is an irreducible component. Since $V_i'\subset X$ is open, we have that if $Z\cap V_i'$ is non-empty, then it is an irreducible component of $V_i'$ (whence $V_i'\subset Z$; thus, $V_i=Z$). Therefore, either there is $i\in I$ such that $V_i'$ meets $Z$ (in which case $V_i=Z$), or $Z\subset H$ (whence $Z=H$). $\square$
