From Munkres' Topology, I get this question. A hint suggests us to use a result from a previous subquestion. But it seems that I don't need to use the previous result to prove this. Can someone help me to see if I can do that? The picture is the exercise that I'm working on. Following that is a proof for $(c)$.
Proof： $$\begin{align*} ||x+y||^2 &= (x+y)(x+y)\\ &=(x+y)x+(x+y)y\\ &=xx+yx+xy+yy\\ &=||x||^2+2(xy)+||y||^2\\ & \leq ||x||^2+2||x||||y||+||y||^2\\ &=(||x||+||y||)^2. \end{align*}$$ Since $||x+y||$ and $(||x||+||y||)$ in the set of nonnegative real numbers, we have $$||x+y||\leq||x||+||y||.$$