Suppose that $X_{t} = \sup(n:\xi_{1}+\ldots+\xi_{n}\leq t)$, $\{\xi_{i}\}$ are i.i.d and $\xi_{i}\sim \mathrm{Exp}(\lambda)$. What is the limit (a.s.) of $X_{t}/t$ (as $t$ approaches $+\infty$)?
I don't have any experience in such problems (regarding convergence of random variables) whatsoever, but the first thing that comes to my mind is the law of large numbers. We can conclude that (!) $$\dfrac{\xi_{1}+\ldots+\xi_{n}}{n} \stackrel{\mathrm{a.s.}}{\to} 1/\lambda$$ Well, we can extract a subsequence of $n$, such that $n = X_{t}$ and so we get $$\dfrac{X_{t}}{\xi_{1}+\ldots+ \xi_{n}}\stackrel{\mathrm{a.s.}}{\to} \lambda$$ One can notice that $\xi_{1}+\ldots+\xi_{n}$ (for $n = X_{t}$) is "close" to $t$. Let's return to (!). We can observe that $$\dfrac{\xi_{1}+\ldots+\xi_{n}}{X_{t}}\leq\dfrac{t}{X_{t}}<\dfrac{\xi_{1}+\ldots+\xi_{n+1}}{X_{t}}$$ The last thing to notice is that $\xi_{n+1}/n$ converges to $0$ almost everywhere (is that true?), so we get the desired.
Is that a proof?
The idea is correct. Specifically, since $X_t\to\infty$ almost-surely as $t\to\infty$ and $(\xi_1+\dots+\xi_n)/n\to \lambda^{-1}$ almost-surely, we get $$\frac{\xi_1+\dots+\xi_{X_t}}{X_t}\to \lambda^{-1}\hspace{1cm}\text{almost-surely}.$$ Now you can use the bound you wanted to use: Since by definition of $X_t$ we have $\xi_1+\dots+\xi_{X_t}\leq t\leq \xi_1+\dots+\xi_{X_t}+\xi_{X_t+1}$ we obtain the sandwich $$\frac{\xi_1+\dots+\xi_{X_t}}{X_t}\leq \frac{t}{X_t}\leq \frac{\xi_1+\dots+\xi_{X_t}+\xi_{X_t+1}}{X_t}$$ which proves the result.