Basic question on Open Sets I'm trying to prove the following:

Let $X$ be a topological space, and $U \subset X$ be an open set, and $C \subset X$. Then, every open in $C \cap U$ is open in $C$.

My proof: $U \cap C$ is open in $C$, directly by definition of the subspace topology. Thus for every open $A \subset (U \cap C)$, there $\exists B \subset C$ such that $B \cap (U \cap C) = A$. So $A$ is open in $C$ as an intersection of two open sets.
Firstly, is this proof correct?
Also, apparently there's another proof that goes: Every open in $U$ is open in $X$, as $U$ is open. It follows that every open in $C \cap U$ is open in $C$. I don't understand how it follows; could someone demistify this (in a more step-by-step manner)? Thanks.
 A: If $A$ is an open set of the subspace $U \cap C$, then $A = (U\cap C)\cap B$, for some open set $B$ in $X$. $B$ need not be a subset of $C$.
But from there the conclusion holds, because $B \cap U$ os open (they are both open in $X$), hence $A$ is open in the subspace topology of $C$.
The second argument is really very similar: By saying that every open set in the subspace topology on $U$ is open in $X$, we are really saying that $U \cap O$ is open in $X$, for every open $O$ in $X$. Then consider an open set, $H$, of $C \cap U$: $H = (C \cap U) \cap O$, for some open $O$ in $X$. But $U \cap O$ is an open set of $U$ and hence an open set of $X$. Hence $H$ must be open in $C$, by the definition of the subspace topology on $C$.
Unless I'm missing something, this second argument is essentially redundant. It's just asserting the fact that $U \cap O$ is open in $X$. But this is immediate from the openness of both $U$ and $O$ anyway; you don't really need to invoke the subspace of $U$ to achieve this.
