I was able to solve part A by utilizing the definition of continuity as well as a clever application of the reverse triangle inequality. The notation for parts b and c are really throwing me off, I think for b it suffices to show that the distance between X1 and Xn is the same as the distance between f1(v) and f2(v). Any help would be appreciated, instead of voting to close my question...
For item $(b)$ you have to observe that $f_{j} = pr_{j}\circ f$ then assuming $f$ continuous you'll have that $f_{j}$ is continuous for every $j$. On the other hand, if you consider $f_{j}$ continuous for every $j$ you have that for every $\epsilon > 0$ there exists $\delta_{j} > 0$ such that $\|x - y\|_{V} < \delta_{j}$ implies that $\|f_{j}(x) - f_{j}(y)\|_{j} < \frac{\epsilon}{n}$, then taking $\delta = \min\{\delta_{j}, j= 1,..., n\} > 0$ you have that if $\|x - y\|_{V} < \delta$, $\|f(x) - f(y)\| = \|f_{1}(x) - f_{1}(y)\|_{1} + ... + \|f_{n}(x) - f_{n}(y)\|_{n} < \epsilon.$ Item $(c)$ is analogous since $f(x)g(x) - f(y)g(y) = (f(x) - f(y))g(x) + f(y)(g(x) - g(y))$.