Let $x(w, q)$ denote the solution to the cost minimization problem :
\begin{eqnarray*} \min_{x} & \ w\cdot x \\ \text{s.t.} & \ \ f(x) \geq q \end{eqnarray*}
where $f$ is the production function.
Since $x(w, q)$ minimizes cost at $(w, q)$, following holds for all $w$ and for all $q$ :
\begin{eqnarray*} w\cdot x(w, q) \leq w\cdot x(w', q) \ \ \ \forall w' \end{eqnarray*}
Also we know that cost function is cost of cost minimizing input choice :
\begin{eqnarray*} C(w, q) = w\cdot x(w, q) \end{eqnarray*}
First, we'll show that $C$ is concave in $w$. Consider any arbitrary $w'$, $w''$ and an arbitrary $\lambda \in [0, 1]$.
\begin{eqnarray*} C(\lambda w' + (1-\lambda) w'', q) & = & (\lambda w' + (1-\lambda) w'')\cdot x(\lambda w' + (1-\lambda) w'', q) \\ & = & \lambda w' \cdot x(\lambda w' + (1-\lambda) w'', q) + (1-\lambda) w''\cdot x(\lambda w' + (1-\lambda) w'', q) \\ & \geq & \lambda w' \cdot x(w', q) + (1-\lambda) w''\cdot x(w'', q) \\ & = & \lambda C(w' , q) + (1-\lambda) C(w'', q) \end{eqnarray*}
Therefore, $C$ is concave in $w$.
To show that input demand is downward sloping, consider any arbitrary $w'$, and $w''$,
\begin{eqnarray*} w'\cdot x(w', q) & \leq & w'\cdot x(w'', q) \\ w''\cdot x(w'', q) & \leq & w''\cdot x(w', q) \end{eqnarray*}
Adding them we get
\begin{eqnarray*} (w'-w'')\cdot (x(w', q) - x(w'', q)) & \leq & 0 \end{eqnarray*}
We now consider a special case where $w'$ and $w''$ differ only in the $j$th input's price. Consequently,
\begin{eqnarray*} (w'-w'')\cdot (x(w', q) - x(w'', q)) = (w_j'-w_j'') (x_j(w', q) - x_j(w'', q)) & \leq & 0 \end{eqnarray*}
Therefore, demand for $j$th input is inversely related to its price.