= Solution
For $x>0$, reflect a Brownian path after its first hitting of $x$. The <Strong Markov property> and symmetry preserve its law and exchange the events $\{S_1\geq x,\ B_1<x\}$ and $\{B_1>x\}$. The terminal equality event has <probability> zero, so the <Brownian reflection principle> gives
$$
\mathbb P(S_1\geq x)=2\mathbb P(B_1\geq x),\qquad
\mathbb P(S_1\leq x)=2\Phi(x)-1.
$$
Continuity of this distribution also gives no atom at zero. Therefore
$$
\boxed{\lim_{x\downarrow0}\frac{\mathbb P(S_1\leq x)}x=2\Phi'(0)=\sqrt{\frac2\pi}.}
$$
A useful global bound follows from the maximum of the standard <normal distribution> density:
$$
\mathbb P(S_1\leq x)=2\int_0^x\frac{e^{-u^2/2}}{\sqrt{2\pi}}\,du\leq\sqrt{\frac2\pi}\,x\qquad(x\geq0).
$$
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