Isometry Preserves Straight Lines Proof?

I'm asked to prove as a corollary to: the image of the line segment PQ under F is a line segment between F(P) and F(Q), that an isometry preserves straight lines.

Here is the proof I've written down:

Let F be an isometry.

Let F(P) be denoted by P'.

Let P,Q be arbitrary points on a line L. Let X be a point on the line segment PQ.

Since F preserves distances we know that: d(P,X)=d(P',X') and d(Q,X)=d(Q',X').

We have d(P,Q)=d(P,X)+d(X,Q).

By assumption of F preserving distance we also have d(P',Q')=d(P',X')+d(X',Q').

Thus the image of segment PQ is contained in its image, segment P'Q'; but since P,Q are arbitrary, the segment PQ may be considered indefinitely long. Thus the line L is mapped to a straight line L', since the arbitrarily large segments composing that line are straight.

Is it correct? Am I writing too much? Is there a better way? Am I missing a second part? ie. to prove to sets equal you must show each belongs to the other.

How to Expand Proof to be for a line?

How can I expand it to include a line if all I know is that a line is determined by two points? It seems to me that I can only deal with segments, or rays I guess. But again, it seems to me to boil down to proving that arbitrary segments on the oppositely directed rays, say XP and XQ, are straight. hmm...?

Plato, from your reply it seems like there is some other way to prove the lines are straight. I'm guessing by taking 3 arbitrary collinear points and proving as I have already done. Is that right? How do the other two cases determine the entire line?

Thanks,

Ultros