Let $\sigma:\mathbb R_+\times\mathbb R\to [1,2]$ be measurable. Consider the SDE $dX_t = \sigma(t,X_t)dW_t$, where $X_0>0$ is independent of Brownian motion $(W_t)_{t\ge 0}$. For every $T>0$ and $R>0$, can we always show
$$\mathbb P[\inf_{0\le t\le T}X_t>0]>0 \mbox{ and } \mathbb P[X_T>R]>0?$$
Here we assume the existence of the solution to the above SDE.
The second inequality is also true. Note that $\mathbb P(X_T>R)>0$ iff $\mathbb E\big((X_T-R)^+\big)>0$. Following D-S theorem (see e.g. https://almostsuremath.com/2010/04/20/time-changed-brownian-motion/), there exists a Brownian motion $B$ with respect to some filtration ${\{\mathcal{G}_t\}_{t\ge 0}}$ s.t. for each ${t\ge 0}$, $\omega\mapsto\langle X\rangle_t(\omega)$ is a ${\mathcal{G}_\cdot}-$stopping time and ${X_t-X_0=B_{\langle X\rangle_t}}$. By assumption, one has $t\le\langle X\rangle_t \le 2t$ for all $t>0$, which implies by Jensen's inequality and Brownian motion's properties
$$\mathbb E\big((X_T-R)^+\big) = \mathbb E\big((X_0+B_{\langle X\rangle_t}-R)^+\big) \ge \mathbb E\big((X_0+B_T-R)^+\big)>0,\quad \forall T>0,$$
as the function $(\cdot-R)^+$ is convex. This allows to conclude.
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 $$\mathbb P[\inf_{0\le t\le T}X_t> 0]\ge \mathbb P[|W_1|<X_0/\sqrt{2T}]>0.$$
