There have been a lot of elegant and precise examples in response to this question, but I want to show that actually, you don't need elegance or precision: there are loads of possible answers to this kind of question.
Start with your candidate number, say $\sqrt 2$. Choose some range about it, say $\sqrt{2} \pm 1$. Now, that's a reasonable chunk of the number line, so you must be able to find a rational number in it, because rational numbers are packed in everywhere. $3/2$ will do, that's the first element of your sequence. Now choose a smaller range, say $\sqrt{2} \pm 1/2$. It turns out $3/2$ fits in this one too, so let's pick that one again. Now choose $7638$. That's nowhere near where you want to be, but it actually doesn't matter – if you only make finitely many "mistakes" like that, it doesn't make the slightest bit of difference to where you converge.
Okay, now let's pick some rational in the range $1$ to $1.5$, because that's a smaller range than before that still contains $\sqrt{2}$. Let's pick $1.23523$ (notice this is actually further away than $3/2$ was – this doesn't matter). Narrow the range to $1.25$ to $1.45$ (you can check this range is still ok by squaring both ends and checking the lower one falls below $2$ and the upper one above it). Pick another rational. Narrow again, and pick again. Keep going forever.
As long as the width of the ranges you allow eventually approach zero, the rational sequence you pick will eventually converge to $\sqrt 2$.
Here's another approach: pick a sequence that converges to $0$. Add $\sqrt{2}$ to every element. Now your sequence converges to $\sqrt{2}$, but it might not have rational elements. However, you can fix it: as before, rationals are packed in everywhere, so you ought to be able to move each element of your sequence just a tiny amount – say, less than its distance to $\sqrt{2}$, so it ends up less than twice as far away – and hit a rational number nearby. Now all your sequence members are rational, and you're done.