= Solution
Set $g_\epsilon(x)=x/\sqrt{\epsilon^2+x^2}$ and $g(x)=\operatorname{sgn}(x)$, with $g(0)=0$. For every $s>0$, the <normal distribution> of $B_s$ has no atom at zero, so $g_\epsilon(B_s)\to g(B_s)$ almost surely. Since $|g_\epsilon-g|\leq2$, the <dominated convergence theorem> and <Tonelli theorem> give
$$
\mathbb E\int_0^T|g_\epsilon(B_s)-g(B_s)|^2ds\longrightarrow0.
$$
The <Itô isometry> followed by the <Doob L2 maximal inequality> now yields
$$
\mathbb E\sup_{t\leq T}\left|
\int_0^t\bigl(g_\epsilon(B_s)-g(B_s)\bigr)dB_s
\right|^2
\leq4\mathbb E\int_0^T|g_\epsilon(B_s)-g(B_s)|^2ds\longrightarrow0.
$$
Thus the stochastic integrals converge ucp to $\int_0^t\operatorname{sgn}(B_s)dB_s$.
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