The word onto is important here. Embed the x axis into the plane, and the open interval (0,1) maps to a set that is neither open nor closed in the plane. This is not a bicontinuous map.
If f is bicontinuous then an open ball centered at the origin maps to an open set containing the origin. This means the image encloses the origin, i.e. it contains a ball about the origin. conversely, assume every open ball centered at the origin has an image that encloses the origin. The image of an open ball at x is, by linearity, f(x) plus the image of the same open ball at 0. Thus the image of the open ball at x includes an open ball about f(x). Apply this to every x in an open set in the domain, and the image is open in the range. This makes f bicontinuous. We only need prove the open ball property.
Cover the domain with open balls centered at the origin, having radius k, for all positive integers k. Let W be the image of the open unit ball in the range. Let kW be the image of the ball of radius k. Note that kW is in fact all the points of W multiplied by k, which sort of justifies my notation.
The images kW, for all k, cover the range.
If W′ is the closure of W, verify that k times W′ is the closure of kW. Since multiplication by k implements a homeomorphism, a point p is in an open set missing W iff kp is in an open set missing kW, hence k×W′ = (kW)′. Since there is no ambiguity, I'll just write kW′.
Suppose W is a nowhere dense set. This means every open ball contains another open ball that misses W. Given an integer k, consider kW, and an open ball. Contract everything by k, and the open ball pulls back to a smaller open ball, nearer the origin. This open ball contains some open ball that misses W, and when we expand by k again we find an open ball that misses kW. Therefore kW is a nowhere dense set.
The range is now the countable union of nowhere dense sets, and that makes it first category. However, a complete metric space is second category. This is a contradiction, hence W is not a nowhere dense set.
If W′ is the closure of W, then W′ contains an open ball of radius r, centered at c. Suppose c is nonzero. If W′ is translated by -c, the open ball will move to the origin. Let's see if we can make this happen. Let d be a preimage of c; hence |d| < 1. Within the ball of radius 2, it is possible to translate the ball of radius 1, so that d moves back to the origin. Since 2W′ contains W′, and translates thereof, it contains an open ball of radius r at the origin. Thus 2W′ encloses the origin. Scale this by any real number, and the closure of the image of every open ball at 0 encloses 0.
Let y0 be any point in the range with |y| < r/2. Thus y0 is contained in W′. Let y1 be a point in W that is within r/4 of y0, and let x1 be a preimage of y1. Thus |x1| < 1.
Let d1 = y0-y1. The point d0 is in ½W′, or perhaps an even smaller image; ratchet down as necessary. Since d1 is in the closure of ½W, find y2 within r/8 of d1, and let f(x2) = y2.
Let d2 = d1-y2, which is y0 - (y1+y2). Since d2 is in ¼W′, find y3 within r/16 of d2, and f(x3) = y3. Then set d3 = d2-y3, find y4 and x4, and so on. continue this process, building a sequence x in the domain and y in the range. The x sequence approaches 0 geometrically. The partial sums of x form a cauchy sequence, with a limit that I will call x0. Meanwhile the y sequence approaches y0. Since f is continuous, f(x0) = y0.
How far is x0 from the origin? Since x1 is in the preimage of W, it has norm at most 1. Similarly, x2 has norm at most ½, x3 has norm at most ¼, and so on. Thus |x| < 2. This holds for all y with |y| < r/2, hence the open ball of radius r/2 is contained in 2W. In other words, W encloses the origin. This proves f is bicontinuous.
If f is injective it is a homeomorphism.