I know this proof is probably super easy but I'm really stuck. I don't want someone to solve it for me, I just want a hint. One way is trivial: suppose f continuous. [0,1] compact and the continuous image of a compact space is compact so f([0,1]) is compact Now the other direction... here's some basic facts i have to work with: 1) R is metrizable 2) R is hausdorf 3) R is regular 4) R is normal 5) f([0,1]) is compact (by assumption) (so it follows that every open cover has a finite subcover) 6) f([0,1]) is closed (follows from 5 because R is Hausdorf and compact subset of Hausdorf space is closed) 7) R is NOT compact, but it's connected 8) [0,1] is connected and compact (every open cover has finite subcover, and no separate of [0,1]) I don't know, everything I try I Just get no where. Can someone give a hint on how/where to start? In showing f is continuous should I try the approach 1) pre-image under f of every open set U in f([0,1]) is open in [0,1] ? 2) pre-image under f of every closed set U in f([0,1]) is closed in [0,1] ? 3) maybe try a contradiction? Assume f is not continuous, which means there exists an open set U in f([0,1]) such that it's pre image is not open, etc.