Flow of a vector field cannot reach a singular point in a finite amount of time? $\newcommand{\R}{\mathbb{R}}$
Let $M$ be a smooth manifold. Let $X \in \Gamma(TM)$ be a smooth vector field, and $\gamma$ some non-constant integral curve of $X$. Let $p \in M$ be a singular point of $X$. ($X(p)=0$).
Questions:
(1) How to prove $\gamma$ cannot approach arbitrarily close to $p$ in a finite amount of time.
More precisely, I want to show the following situation is impossible:
$\gamma:(0,a) \to M \, , \, \lim_{t\to a} \gamma(t)=p$.
(2) What is the minimal regularity required from $X$ for this result to hold. (See a counter example where $X$ is only continuous below).

A Possible approach for proving (1):
A non constant integral curve is an immersion.* 
If we can define $\gamma(a)=p$, and extend $\gamma$ smoothly beyond $t=a$, we could get a contradiction, since $\gamma$ is non constant , and $\dot \gamma(a)=X(p)=0$ so it's not an immersion either. 
The problem is that I do not know how to show $\gamma$ can be extended smoothly to an open neighbourhood of $a$.

A Counter example where $X$ is continuous but not smooth:
$M=\R, X(t)=2\sqrt{|t|}$. Look at $\gamma:(0,1) \to \R, \gamma(t)=t^2$.
$\dot \gamma(t)=2t=2\sqrt{t^2}=X(t^2)=X\big(\gamma(t)\big)$. 
Here $\gamma$ flows out from a singularity, and not towards it, but this changes nothing since we can always change it's direction in time, and obtain an integral curve $\alpha(t)=\gamma(1-t)$ of the vector field $(-X)$.
Note that $\gamma$ is smooth, though $X$ is not differentiable at the singular point $0$.

*See proposition 9.21, Introduction to smooth manifols, by Lee.
 A: I think the relevant issue is the uniqueness of integral curves. In your counterexample, there are two two integral curves for $X$ starting at $t = 0$, the constant integral curve and $\gamma(t) = t^2$. 
Let $\gamma \colon (0, a) \rightarrow M$ be an integral curve of $X$ satisfying $\lim_{t \to a} \gamma(t) = p$ with $X(p) = 0$ and $a < \infty$.
Define $\tilde{\gamma} \colon (0, \infty) \rightarrow M$ by
$$ \tilde{\gamma}(t) := \begin{cases} \gamma(t) & 0 < t < a, \\ p & a \leq t < \infty. \end{cases} $$
Then $\gamma$ is continuous at $t = a$, and 
$$ \lim_{x \to a^{-}} \dot{\tilde{\gamma}}(t) = \lim_{x \to a^{-}} X(\gamma(t)) = X(p) = 0 = \lim_{x \to a^{+}} \dot{\tilde{\gamma}}(t) $$
so $\tilde{\gamma}$ is also differentiable at $t = a$. We have that $\tilde{\gamma}$ is a $C^1$ integral curve of $X$ satisfying $\tilde{\gamma}(a) = p$ and by uniqueness $\tilde{\gamma}$ is constant and so is $\gamma$.
The proof works as long as the uniqueness of integral curves work so I think it is enough to require that $X$ is $C^1$ or even locally-Lipschitz $C^0$.
