Map from 2-sphere into $(\mathbb R^3, |\cdot|)$ Can you help me with this?

Let $S^2 := \{x\in \mathbb R^3:||x||_2 = 1\} \subset (\mathbb R^3, ||\cdot||_2)$  and $T:S^2 \to (\mathbb R, |\cdot|)$ be a continuous map. Since $S^2$ is compact, $T$ attains its maximum $T_{max}$ and its minimum $T_{min}$.
  Is there a value $T\in]T_{min},T_{max}[$ that is assumed at only one $x_0\in S^2$?   

I think I have to use the fact that the image of a connected space under a continuous function is connected again, but I don't know how. Can you help?

This is the answer by @Hmm. expanded.
Suppose there was such a value. Then, there would exist a induced map $\widetilde{T}: S^2 \backslash \{x_0\} \to [T_{min},T[ \cup ]T,T_{max}]$, obtained by restricting the domain first, and then the image. This induced map is continuous by simple properties of continuity. But $S^2\backslash \{x_0\}$ is connected, since it is homeomorphic to $\mathbb{R^2}$. Therefore, the whole image must be inside either $[T_{min}, T[$ or $]T,T_{max}]$, which is absurd, since there is a point which is taken to $T_{min}$ and one which is taken to $T_{max}$.
