In my previous post Why can't you consider coideal generated by sets, where consequently i've shown why intersection of coideals need not to be a coideal, i said that
... the nonempty family $\{ I \supseteq X \colon I \text{ coideal of } C\}$ have minimal elements ...
I shall proof this affirmation for $k$-coalgebras ($k$ a field) bellow.
After de proof, we can think of coideals in the same way as prime ideals in noncommutative rings. In general the intersection of those ideals isn't a prime ideal, but if they're chained one can prove that it's intersection in indeed a prime ideal. The definition of prime ideal is analogous to (1)+(2) below exchanging sum for product, and the proof follows the argument in (4) for ideals contained in a chain intersection.
Considering $C$ a coalgebra over a field is essential here, cause i'll need a very specific basis for $C \otimes C$.
Let $I_1 \supseteq I_2 \supseteq \dots$ be a chain of coideals of $C$. By definition each $I_n$ is a subspace of $C$, thus we can find $W_n$ such that $I_n = W_n \oplus I_{n-1}$ for every $n > 0$ and consequently $I_1 = \bigoplus_{n>0} W_n$. Fix a basis $B'_n$ for $W_n$, then $B_n = \cup_{i \geq n} B'_i$ is a basis for $I_n$ and $B_1 \supseteq B_2 \supseteq \dots$. Extend $B_1$ to a basis $B$ of $C$, then $B^2 := \{ v_i \otimes v_j \colon v_i \in B_i,v_j \in B \}$ is a basis for $C \otimes C$.
Let $x \in \bigcap_{j>0} I_j =: I$, then $\Delta(x) \in \Delta(\bigcap_{j>0} I_j) \subseteq \bigcap_{j>0} \Delta(I_j) \subseteq \bigcap_{j>0} (C \otimes I_j + I_j \otimes C)$. Write $\Delta(x) = \sum_{i,j} \alpha_{ij}v_i \otimes v_j$, so...
(1) if $v_i \otimes v_j \notin C \otimes I_n + I_n \otimes C$ for some $n>0$, $\alpha_{ij} = 0$.
(2) if $\alpha_{ij} \neq 0$ then for every $n>0$, $v_i \otimes v_j \in C \otimes I_n$ or $v_i \otimes v_j \in I_n \otimes C$.
(3) if $\alpha_{ij} \neq 0$ and $v_i \otimes v_j \in C \otimes I_n$ for every $n>0$ then $v_i \otimes v_j \in C \otimes I$.
(4) if $\alpha_{ij} \neq 0$ and $v_i \otimes v_j \notin C \otimes I_n$ for some $n>0$ then $v_i \otimes v_j \in I \otimes C$.
By proving (1)-(4) we can rearrange the summands of $\Delta(x)$ to get $$ \Delta(x) = \sum_{i,j} \alpha_{ij} v_i \otimes v_j \overset{2+3}{=} \sum \alpha_{ij}v_i \otimes v_j + \sum \alpha_{rs}v_r \otimes v_s \in C \otimes I + I \otimes C $$ and as $\epsilon(x) \in \epsilon(I_1) = 0$ it follows that $I = \bigcap_{n>0} I_n$ is a coideal of $C$.
(1) by $(C \otimes I_n) \cap (I_n \cap C) = I_n \cap I_n$, a basis for $C \otimes I_n + I_n \otimes C$ is given by the disjoint union $$ B' = \{ v_i \otimes v_j \colon v_i \in B \setminus B_n, v_j \in B_n \} \cup \{ v_i \otimes v_j \colon v_i,v_j \in B_n \} \cup \{ v_i \otimes v_j \colon v_i \in B_n, v_j \in B \setminus B_n \} \subseteq B $$ the two on the left generates $C \otimes I_n$ while the two on the right generate $I_n \otimes C$, thus the union generates the sum, and $B'$ is linearly independent as it is a subset of a basis. Write $\Delta(x) = \sum \beta_{ij} v_i \otimes v_j$ with $v_i \otimes v_j \in B'$, then $$ 0 = \sum_{v_i \otimes v_j \in B^2} \alpha_{ij} v_i \otimes v_j - \sum_{v_i \otimes v_j \in B'} \beta_{ij} v_i \otimes v_j = \sum_{v_i \otimes v_j \in B'} (\alpha_{ij}-\beta_{ij})v_i \otimes v_j + \sum_{v_i \otimes v_j \in B^2 \setminus B'} \alpha_{ij}v_i \otimes v_j $$ and by uniqueness of writing $\alpha_{ij} = 0$ for all $(i,j)$ such that $v_i \otimes v_j \in B^2 \setminus B' = B^2 \cap (C \otimes I_n + I_n \otimes C)^c$.
(2) in view of (1), if $\alpha_{ij} \neq 0$ then $v_i \otimes v_j \in (C \otimes I_n + I_n \otimes C) \cap B = B'$, so $v_i \otimes v_j \in \{ v_r \otimes v_s \colon v_r \in B, v_s \in B_n \}$ implying $v_j \in C \otimes I_n$ or $v_i \otimes v_j \in \{ v_r \otimes v_s \colon v_r \in B, v_s \in B_n \}$ implying $v_i \otimes v_j \in I_n \otimes C$.
(3) let $BB_n = \{ v_i \otimes v_j \colon v_i \in B, v_j \in B_n \}$, then $BB_n$ is a basis for $C \otimes I_n$ so $$ v_i \otimes v_j \in \bigcap(C \otimes I_n) \cap B^2 = \bigcap((C \otimes I_n) \cap B^2) = \bigcap BB_n = \{ v_r \otimes v_s \colon v_r \in B, v_s \in B_n, \forall n > 0 \} $$ the result follows from $v_j \in \bigcap B_n \subseteq \bigcap I_n = I$.
(4) fix $n_0>0$ such that $v_i \otimes v_j \notin C \otimes I_{n_0}$ so $v_i \otimes v_j \notin C \otimes I_n$ for all $n \geq n_0$ because $C \otimes I_{n_0} \supseteq C \otimes I_n$. From (2) it follows that $v_i \otimes v_j \in I_n \otimes C$ forall $n \geq n_0$ and thus $v_i \otimes v_j \in I_{n_0} \otimes C \subseteq I_{n_0-1} \otimes C \subseteq \dots \subseteq I_1 \otimes C$, ie, $v_i \otimes v_j \in I_n \otimes C$ for all $n>0$. Following the argument in (3) we prove that $v_i \otimes v_j \in I \otimes C$.
From this $I = \bigcap_{n>0} I_n$ is a coideal and we can apply Zorn's lemma to find minimal elements in the family $\{ I \supseteq X \colon I \text{ coideal of } C \}$ as long as $X \subseteq ker(\epsilon)$.