Since $x_i \leq c$,
$\displaystyle \sum_i x_i^2 = \sum_i x_i\cdot x_i \leq \sum_i c\cdot x_i = cn\bar{x}.$
Note also that $0 \leq \bar{x} \leq c$. Then,
$$
\begin{align*}
n\cdot \text{var}(\mathbf{x}) &= \sum_i (x_i - \bar{x})^2= \sum_i x_i^2 - 2x_i\bar{x} + \bar{x}^2\\
&= \sum_i x_i^2 - 2\bar{x}\sum_i x_i + n\bar{x}^2= \sum_i x_i^2 - n\bar{x}^2\\
&\leq cn\bar{x} - n\bar{x}^2 = n\bar{x}(c-\bar{x})
\end{align*}
$$
and thus $$\text{var}(\mathbf{x}) \leq \bar{x}(c-\bar{x}) \leq \frac{c^2}{4}.$$
Added note: (second edit)
The result $\text{var}(X) \leq \frac{c^2}{4}$ also applies to random variables
taking on values in $[0,c]$, and, as my first comment on the question says, putting half the mass at $0$ and the other half at $c$ gives the maximal variance of $c^2/4$. For the vector $\mathbf x$, if $n$ is even, the maximal variance $c^2/4$ occurs when $n/2$ of the $x_i$ have value $0$ and the
rest have value $c$.
Someone else posted an answer -- it has since been deleted -- which said the
same thing and added that if $n$ is odd, the variance is maximized when
$(n+1)/2$ of the $x_i$ have value $0$ and $(n-1)/2$ have value $c$,
or vice versa. This gives a variance of $(c^2/4)\cdot(n^2-1)/n^2$ which
is slightly smaller than $c^2/4$. Putting the "extra" point at $c/2$
instead of at an endpoint gives a slightly smaller variance of
$(c^2/4)\cdot(n-1)/n$, but both choices have variance approaching
$c^2/4$ asymptotically as $n \to \infty$.