Another one, by contradiction (which is essentially the same as the previous one) is: if x_n tends to x and x_n tends to y. If x=/=y, then let e=|x-y|/3, then for all n sufficently large we have
|x_n - x| < e , ie all the x_n are in the interval (x-e,x+e) for n >N
|x_n - y| <e, ie all the x_n are in the interval (y-e,y+e) for n>N
but the two intervals (x-e,x+e) and (y-e,y+e) are disjoint so it is impossible for all the x_n for n>N to be in two disjoint intervals. a contradiction. the only assumption was that x=/=y, so it must be that x=y.
Personally I prefer this one (even though it is an unnecessary proof by contradiction) because it makes you think visually (or geometrically) about what it means for a seqqunce to converge: for all open intervals about x (or open "balls" in higher dimensions) that if we throw away some finite number of terms at the start of the sequence then all of the ones that are left are inside the inteval or ball. it gets awayy from the epsilon delta arguments, and that has to be a good thing.