I've proven DC, namely:
Let $R\subset a \times a$ so that for each $x\in a$ there is some $y\in a$ with $(y,x)\in R$. Then there is some function $f: \omega \rightarrow a$ so that for any $n\in \omega$ we have $(f(n+1), f(n))\in R$.
Now, I want to use this fact to prove that $R\subset B\times B$ is well-founded iff there is no function $f: \omega \rightarrow B$ such that $(f(n+1), f(n))\in R$ for each $n\in \omega$. (By the way, well-foundedness here means each nonempty subset of $B$ has an $R$-least element.)
However, I haven't been able to work out the proof. Am I wrong to use DC for this proof? If not, how would it go?
I am guessing that you are trying to prove that well-foundedness equivalent to the Principle of Dependent Choice.
So assume that $R$ is a well-founded relation on some non-empty set $B$. If there is an infinite decreasing chain, then clearly $R$ is not well-founded.
The other direction, however, requires that we use Dependent Choice. Suppose that $R$ is not well-founded. Let $A$ be a set without a minimal element, witnessing that $R$ is not well-founded. Define $X$ to be the set of finite sequences of $R$-decreasing sequences from $X$, with $S$ being the end-extension relation on $X$.
Now, using Dependent Choice, we can find an $S$-sequence in $X$. The $S$-sequence's limit is a function from $\omega$ to $B$ as required.