Let $\tau = \inf\{ t>0 : W_t = 1 \}$.  The conjecture is **true** and the essence of the proof outlined below appears to be the following peculiar property of the hitting time $\tau$: $$
\lim_{\epsilon \searrow 0} P_{\epsilon}[\tau > \epsilon-\epsilon^{3/2}] = 1 \;. $$ 
(This can be computed directly using the fact that distribution of $\tau$ is [inverse gamma][1] with parameters $1/2$ and $1/2$.) In order to leverage this property, one must carefully split up $e_t:= X_t - Y_{t/\epsilon}$ as outlined below.  


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> **Theorem** It holds that:  $\lim_{\epsilon \searrow 0}  E[ \sup_{0 \le t \le \epsilon} |e_t|^2 ] = 0 \;.$

*Proof*. By Itô's formula \begin{align*}
& |e_t|^2 = \mathrm{I} +  \mathrm{II} +  \mathrm{III} \quad \text{where} \\
&  \mathrm{I}:= \frac{2}{\epsilon} \int_0^t e_s (\sigma(X_s) - \sigma(Y_{s/\epsilon})) ds \;, \\
&  \mathrm{II}:= \frac{2}{\epsilon} \int_0^t e_s \sigma(X_s) (\epsilon dW_s - ds) \;, \\
&  \mathrm{III}:= \int_0^t \sigma(X_s)^2 ds \;. 
\end{align*}
The proof shows that $$
E_{\epsilon} [ \sup_{0 \le t \le T} |e_t|^2 ] \le C_1(\epsilon) + C_2 (\epsilon) \int_0^T E_{\epsilon} [ \sup_{0 \le r \le s} |e_r|^2 ] ds  
$$ where $C_1(\epsilon)$ and $C_2 (\epsilon) $ are non-negative and $\lim_{\epsilon \searrow 0}  C_1(\epsilon) =0$ and $\lim_{\epsilon \searrow 0} C_2(\epsilon) \epsilon = O(1)$.  By Grönwall's inequality, $$
E_{\epsilon} [ \sup_{0 \le t \le \epsilon} |e_t|^2 ] \le C_1(\epsilon) \exp(C_2 (\epsilon) \epsilon) 
$$ as required.  


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*Estimate for $\mathrm{I}$.*

This term exclusively contributes to $C_2(\epsilon)$.
Since $\sigma$ is $L$-Lipschitz for some $L>0$ $$
I \le \frac{2 L}{\epsilon} \int_0^t |e_s|^2 ds 
$$ and thus $$
\sup_{0 \le t \le \epsilon} I \le \frac{2 L}{\epsilon} \int_0^{\epsilon} |e_s|^2 ds \le \frac{2 L}{\epsilon} \int_0^{\epsilon} \sup_{0 \le r \le s} |e_r|^2 ds \;.
$$
Thus, $C_2(\epsilon) = 2 L / \epsilon$.

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*Estimate for $\mathrm{II}$.*

This term contributes to $C_1(\epsilon)$, and here is where we leverage the peculiar property of $\tau$ mentioned in the intro.  
\begin{align*}
& \lim_{\epsilon \searrow 0} E_{\epsilon} [ \sup_{0 \le t \le \epsilon} \left| \mathrm{II} \right| ]  = 
\lim_{\epsilon \searrow 0} E_{\epsilon} [ \sup_{0 \le t \le \epsilon} \left| \mathrm{II} \right| \mathbf{1}_{ \{ \tau > \epsilon - \epsilon^{3/2} \} } ] \\
& \quad = 
\lim_{\epsilon \searrow 0} E [ \sup_{0 \le t \le \epsilon} \left| 2 \int_0^t e_s \sigma(X_s) dW_s \right| ] = 0
\end{align*}
where the second to last step follows because conditioned on the event $(\tau < \epsilon)$, the law of $\epsilon W_s - s$ is equal to the law of a standard Brownian bridge, and in the last step we used [Doob's martingale inequality][2], [Itô isometry][3], and (standard) a priori bounds on $X_t$ and $Y_{t/\epsilon}$ over $(0,\epsilon)$.

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*Estimate for $\mathrm{III}$.*

This term also contributes to $C_1(\epsilon)$.  Noting that $\sigma$ is $L$-Lipschitz, \begin{align*}
E_{\epsilon} [ \sup_{0 \le t \le \epsilon} \mathrm{III} ] &= E_{\epsilon}[ \int_0^{\epsilon}  [ \sigma(X_s)^2 ds ] \\
&\le 2 \epsilon \sigma(0)^2 + 2 L^2 E_{\epsilon}[ \int_0^{\epsilon} |X_s|^2 ds ] \\
&\le 2  \epsilon ( \sigma(0)^2 + L^2 |x_0|^2 ) + 2 L^2 \epsilon \int_0^{\epsilon} \sigma(X_s)^2 ds \\ 
& \quad + 2 L^2 \epsilon \int_0^{\epsilon} X_s \sigma(X_s) dW_s
\end{align*}
and as long as $2 L^2 \epsilon \le 1/2$, it follows that 
$$
E_{\epsilon} [ \sup_{0 \le t \le \epsilon} \mathrm{III} ] \le 4  \epsilon ( \sigma(0)^2 + L^2 |x_0|^2 ) + 4 L^2 \epsilon \int_0^{\epsilon} X_s \sigma(X_s) dW_s
$$ The last term in this expression can be treated in a similar way as the last step in the estimate for $\mathrm{II}$, namely [Doob's martingale inequality][2], Cauchy-Schwarz, [Itô isometry][3], and (standard) a priori bounds on $X_t$ over $(0,\epsilon)$.  In particular, \begin{align*}
\left( E_{\epsilon} [ \sup_{0 \le t \le \epsilon} \left| \int_0^t X_s \sigma(X_s) d W_s \right| ] \right)^2 &\le
E \sup_{0 \le t \le \epsilon} \left| \int_0^t X_s \sigma(X_s) dW_s \right|^2\\
&\le 4 E \left| \int_0^{\epsilon} X_s \sigma(X_s) dW_s \right|^2 \\
&\le 4 E \int_0^{\epsilon} X_s^2 \sigma(X_s)^2 ds \\
&\le 4 \tilde{C}_2 (1+ x_0^4) e^{\tilde{C}_1 \epsilon} \epsilon
\end{align*} 
where in turn we used Cauchy-Schwarz, Doob's martingale inequality with $p=2$, Itô's isometry, and then an a priori bound on the second/fourth moment of $X_t$ over $(0, \epsilon)$.  





  [1]: https://en.wikipedia.org/wiki/Inverse-gamma_distribution
  [2]: https://en.wikipedia.org/wiki/Doob's_martingale_inequality#Further_inequalities
  [3]: https://en.wikipedia.org/wiki/It%C3%B4_isometry