This is a partial answer. Conditionning on $\{X_0=x\}$, one has

$$\mathbb P[\inf_{0\le t\le T}X_t> 0]= \int_{(0,\infty)}\mathbb P[\inf_{0\le t\le T}(X_t-X_0)>-x|X_0=x]\mathbb P[X_0\in dx].$$

As any continuous martigale starting from zero is a time changed Brownian motion, there exists some Brownian motion, denoted by $B$, s.t. $X_t-X_0=B_{\langle X\rangle_t}$, where $\langle X\rangle_t:=\int_0^t\sigma(s,X_s)^2ds \in [t, 2t]$. Thus
\begin{eqnarray}
\mathbb P[\inf_{0\le t\le T}(X_t-X_0)> -x|X_0=x]&=&\mathbb P[\inf_{0\le t\le T}B_{\langle X\rangle_t}> -x|X_0=x]\\
&\ge& \mathbb P[\inf_{0\le t\le 2T}B_t>-x|X_0=x]\\
&=& \mathbb P[|B_1|<x/\sqrt{2T}],
\end{eqnarray}
which yields
$$\mathbb P[\inf_{0\le t\le T}X_t> 0]\ge \mathbb P[|W_1|<X_0/\sqrt{2T}]>0.$$