any continuous function is null homotopic for convex set. Let $X$ be a topological space. and suppose $B$ is a convex subset in $\mathbb{R}^n$. Prove that any continuous map $f: X \rightarrow B$ is null-homotopic.

My strategy is following the defintion of null-homotopic which is homotopic to constant function $c$. 
To prove this find homotopoy $H$ satisfying
\begin{align}
H(t,0) = f(t) \in B \qquad  H(t,1) = c   \in \mathbb{R}^n
\end{align}
 define $H$ as 
\begin{align}
  H(t,s) = (1-s) f(t) + s c\in \mathbb{R}^n
\end{align}
which is continuous  and satisfies 
\begin{align}
  H(t,0) = f(t), \qquad H(t,1) = c
\end{align}

Is this valid approach? I feel uncomfortable since in the above process i never used the concept of "convex" explicitly. 

This is correct.  The place you use convexity is in knowing that $H(t,s)\in B$ for all $(t,s)$ (remember, you need $H$ to be a map $X\times[0,1]\to B$, not just a map to $\mathbb{R}^n$!).  Since $f(t)\in B$ and $c\in B$, $(1-s)f(t)+sc\in B$ by convexity.  You probably want to mention this explicitly when writing up your solution, both to justify the fact that $H(t,s)\in B$ and to show your reader where you are using convexity.
