Given any ring $R$, prove that the intersection of an arbitrary collection of subrings of $R$ is also a subring of $R$.
Here is my attempt:
Suppose $R$ is an arbitrary ring. Let $S_{1},S_{2},...,S_{n}$ be subrings of $R$. Clearly $0_{R}\in\bigcap_{i=1}^{n}S_{i}$. Then if $x,y\in\bigcap_{i=1}^{n}S_{i}$, then $x,y\in S_{1},S_{2},...S_{n}$. Since each $S_{i}$ is closed under subtraction and multiplication, $x-y\in\bigcap_{i=1}^{n}S_{i}$, and $xy\in\bigcap_{i=1}^{n}S_{i}$. Hence, $\bigcap_{i=1}^{n}S_{i}$ is a subring of $R$.
I believe that this is correct, but I was also thinking that since the trivial ring is a subring of $R$, i.e. $\{0_{R}\}\in\bigcap_{i=1}^{n}S_{i}$, then the only elements that could be in the intersection would be $0_{R}$. Any advice?
[I assume your rings are not required to have a unit. If they are, you also need to mention $1_R$ being in all the subrings and thus in their intersection, and there is no "trivial subring".]
This is basically correct, but you have not considered an arbitrary collection of subrings. What if your collection of subrings is infinite? You can fix this by just not saying you have some set $C$ of subrings and not explicitly labelling the elements as $C=\{S_1,\dots,S_n\}$.
That would be the case if you took the intersection of all the subrings of $R$, since $\{0_R\}$ is one such subring. But you aren't necessarily intersecting all the subrings. You are just taking some subrings and intersecting them, so the intersection may be larger. (Also, your notation is incorrect: I think you mean to say $\{0_R\}\in \{S_1,\dots,S_n\}$.)