For each positive integer m, let Um,i be the set of x in S with |fi(x)| ≤ m. Let Um be the intersection over i of Um,i. Since each fi is continuous, each Um,i is closed, hence each Um is closed.
By assumption, each x has an m(x) bounding all the images fi(x), which means each x belongs to some Um, and S is the countable union over Um. By the first category theorem, some Um fails to be nowhere dense. It contains an open set, and the image of this open set is bounded by m.
Perhaps S is not a complete metric space, but the functions in f are all uniformly continuous. These functions can be extended to the completion of S. The uniform boundedness principle applies, And since each fi is the same on S and on the completion of S, we can pull this result back to S. Therefore, some open set in S has an image bounded in R1.