Let $A$ be the event $A = \left\{ \limsup_{T \to \infty}\frac{1}{T}\sum_{t = 0}^{T - 1}X_{t} \ge \frac{1}{2}\right\}$, so that $Y_t = P(A \mid X_1, \dots, X_t)$. We will show that $P(A) = 1$ and hence $Y_t = 1$ almost surely for all $t$.

First, Lévy's zero-one law asserts that $Y_t \to 1_A$ almost surely. Now assume there exists $\epsilon < 1/2$ such that for all $t$, we have $P(X_t = 1 \mid Y_t < \epsilon) > 1-\epsilon$ for all $t$ (I'm guessing that's what you want the quantifiers to be). We can rewrite this as
$$P(X_t = 1, Y_t < \epsilon) > (1-\epsilon) P(Y_t < \epsilon) $$
or equivalently, letting $Z_t = 1_{\{Y_t < \epsilon\}}$,
$$E[X_t Z_t] > (1-\epsilon) E[Z_t].$$
Since $Y_t \to 1_A$ we have $Z_t \to 1_{A^c}$ almost surely, so by dominated convergence the right side converges to $(1-\epsilon)P(A^c)$. On the left side, let us add and subtract $E[X_t 1_{A^c}]$:
$$E[X_t(Z_t - 1_{A^c})] + E[X_t 1_{A^c}] > (1-\epsilon)E[Z_t]$$
Since $Z_t \to 1_{A^c}$ almost surely, and $|X_t| \le 1$, the first term on the left goes to 0. Now let us average over $T$:
$$\frac{1}{T} \sum_{t=1}^T E[X_t(Z_t - 1_{A^c})] + E\left[\frac{1}{T} \sum_{t=1}^T X_t 1_{A^c}\right] > (1-\epsilon) \frac{1}{T} \sum_{t=1}^T E[Z_t].$$
The first term on the left side goes to 0 and the right side goes to $(1-\epsilon) P(A^c)$, so we have
$$(1-\epsilon) P(A^c) \le \limsup_{T \to \infty} E\left[\frac{1}{T} \sum_{t=1}^T X_t 1_{A^c}\right].$$
The quantity inside the expectation is bounded above by 1, so Fatou's lemma gives
$$\limsup_{T \to \infty} E\left[\frac{1}{T} \sum_{t=1}^T X_t 1_{A^c}\right] \le E\left[\limsup_{T \to \infty} \frac{1}{T} \sum_{t=1}^T X_t 1_{A^c}\right].$$
But by definition of $A$, on the event $A^c$ we have $\limsup_{T \to \infty} \frac{1}{T} \sum_{t=1}^T X_t \le \frac{1}{2}$. Therefore
$$(1-\epsilon) P(A^c) \le \frac{1}{2} P(A^c).$$
Since $\epsilon < 1/2$ this implies $P(A^c) = 0$.