Imagine a particle moving through 3 space. Its path is defined by three functions: x(t), y(t), and z(t). These functions are continuous iff the path is continuous with respect to the topology of 3 space. Here is the general idea. If each coordinate stays close to x(p) y(p) z(p) respectively, when t is close to p, then the plot of those coordinates in 3 space is never far from the point x(p),y(p),z(p), which we call f(p). Conversely, if f(t) is always close to f(p) when t is close to p, then the individual coordinates can't be far away from x(p), y(p), and z(p). This is proved in detail in product spaces. The heart of the proof is the triangular inequality. You can't be far away unless one of the coordinates is far away.
As long as the number of dimensions (n) is finite, a vector function is uniform iff all its component functions are uniform. Given ε, find a δ for each component function, keeping that coordinate within ε/n of its target. Use the smallest of these, and the entire function is within ε of its limit. Again, the triangular inequality makes this work.