DFA impossible problem?
Note: I am the user who gave the comment cited in the OP on the CSTheory thread. Also, I am writing this as an answer rather than a comment to JohEker's post as I do not have enough reputation here to comment.



The idea is to exploit that we have a finite, local bound on the difference between zeroes and ones in the input word, i.e. we accept every word where this difference does not exceed 1 at any point. Note that if we drop the requirement that this holds for all prefixes, then the language is not regular (and thus not recognized by a DFA) by the Myhill-Nerode theorem as pointed out by JohEker.

So let $\#0(w)$ and $\#1(w)$ denote the number of ones and zeroes in the word $w$ respectively. To build a DFA from the idea above, we see that a word $w$ should be accepted if every prefix $v$ of $w$ satisfies $|\#1(v) - \#0(v)| \leq 1$, or equivalently $-1 \leq \#1(v) - \#0(v) \leq 1$.
We can keep track of the value $\#1(v) - \#0(v)$ by creating a state for every value of it that we care about and specifying transitions between these states to model how the counter changes when we read a $0$ or a $1$.
Combining this with our previous idea that we only need to know if the value of the counter is between $-1$ and $1$, we create three states $q_{-1}, q_0$ and $q_1$. 

Clearly we want the initial value of the counter to be $0$, so we use $q_0$ as the initial state of our automaton and we transition from $q_k$ to $q_{k+1}$ if we read a $1$ (for $k \in \{-1, 0\}$ -- else we leave the space of values that we are interested in; more on this later) and similarly we transition from $q_k$ to $q_{k-1}$ (for $k \in \{0, 1\}$). As we want to accept the words for which the counter ends up at 0, we also define $q_0$ as the only accepting state.

Now, what happens if the counter does not stay within $\{-1, 0, 1\}$? We want to reject the input.

In our automaton, this corresponds either to reading a $0$ while we are in the state $q_{-1}$ (i.e. we have $\#1(v) - \#0(v) = -1$ that for the input $v$ before reading the $0$ and thus $\#1(v0) - \#0(v0) = -2$) or reading a $1$ while we are in the state $q_1$. So we introduce a new, "rejecting" state $q_s$ (also called a sink-state as the DFA cannot escape from this state once it is reached) which we transition to in case we detect one of the above-described scenarios. This now completes the description of our DFA, which looks like this: