How to show that $S_k = \inf \{t \geq 0 | \|X(t)\| \geq k \} \to \infty$ as $k \to \infty$ a.s.

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stack.exchangers! I am currently working my way through the proof given by Karatzas and Shreve (1988) of the Feynman-Kac Theorem (Theorem 5.7.6). However, I am missing out on the following problem:

In the theorem it is being used that if we assume that the $d$-dimensional process $X$ is a weak solution to the stochastic differential equation

$$X(t)=X(s)+\int_s^t b(u,X(u)) \, du+\int_s^t \sigma(u,X(u)) \, dW(u)$$

where $W(t)$ is a $r$-dimensional Brownian motion and we also assume the drift and diffusion coefficient satisfies the linear growth condition

$$\|b(t,x)\|^2+\|\sigma(t,x)\|^2 \leq K^2(1+\|x\|^2)$$

Then it holds that the stopping time

$$S_k \equiv \inf\{t\geq0,\|X(t)\|\geq k\}\to\infty$$ almost surely as $k\to \infty$.

My idea so far would be to use the definition of a weak solution that states that for a weak solution it holds that $P\left( \int_0^t |b_i (s,X_s)| + \sigma_{ij}^2(s,X_{s}) \, ds < \infty\right)=1$ maybe in combination with the linear growth condition but I have not found the correct way yet.

Any help is welcome, hints, solutions or a reference where i can more about it. Thank you.