= Solution
Use the <Itô formula> up to $T_n$:
$$
X_{t\wedge T_n}^2
=2\int_0^{t\wedge T_n}X_s\sigma(X_s)\,dB_s
+\int_0^{t\wedge T_n}\bigl(2X_sb(X_s)+\sigma^2(X_s)\bigr)ds.
$$
For fixed $n$, the stopped stochastic integrand is bounded on each finite horizon, so its integral has <expectation> zero. The <linear growth condition for an SDE> and $2xb(x)\le x^2+b(x)^2$ give
$$
2xb(x)+\sigma^2(x)\le K+(K+1)x^2.
$$
Writing $u_n(t)=\mathbb EX_{t\wedge T_n}^2$, and bounding the active-time integrand by the stopped square, yields
$$
u_n(t)\le Kt+(K+1)\int_0^tu_n(s)\,ds.
$$
The <Gronwall inequality> gives a bound independent of $n$:
$$
u_n(t)\le\frac{K}{K+1}\bigl(e^{(K+1)t}-1\bigr)=C_t.
$$
On $\{T_n\le t\}$, continuity gives $|X_{t\wedge T_n}|=n$. The <Markov inequality> therefore gives
$$
\boxed{\mathbb P(T_n\le t)\le C_t/n^2\longrightarrow0}.
$$
Since $T_n\uparrow\xi$, the event $\{\xi\le t\}$ is the intersection of these decreasing events and has probability zero. Taking integer $t$ proves $\xi=\infty$ almost surely. This proves global existence with locally, rather than globally, Lipschitz coefficients under the stated growth bound.
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