I want to prove that given a net $S$ and topologies $T$ and $T'$, then $T\subset T'\iff$ when $S$ is convergent in $T'$ is also convergent in $T$.
I'm proceeding this way: First I'd like to show that if $T\subset T'$ then if $S\to x$ in $T'$ it happens in $T$ as well, which means that I must show that for every $U$ that satisfies $x\in U\subset T\exists N$ such that $S(n)\in U'$ when $n>N$. Now, take $U$ as before, since $T'$ is finer than $T$, this means that $\exists U'\subset T'$ such that $x\in U'\subset U$, and since $S$ converge in $T'$ we know that there is $N$ for which $S(n)\in U'$ when $n\geq N$, we conclude by means of the former inclusion that $S(n)\in U$ when $n\geq N$.
Proving the reciprocal: Seems immediate, but is it?. $S\to x$ in $T'$ implies that $S\to x$ in $T$. Let $B,B'$ be basis for $T,T'$ and take $U_B$ such that $x\in U_B\subset B$. Given that $S(n)\in U'_B$ for $n\geq N$ for some $N$ implies that $S(n)\in U_B$, then follows that $U'_B\subset U_b$. Then there is $U'_b$ such that $x\in U'_b\subset U_b$, which means that $T'$ is finer than $T$.
Is this proof right?.