Set of consumption over all period is convex in $\mathbb{R}^T_+$? Today in class, the professor said that the set of all consumption $c(S)$ is non-empty, compact and convex subset of $\mathbb{R}^T_+$. i.e. we know $\sum \limits_{t=1} ^T c_t = S$ where $c_t$ is consumption in period t and $S$ is total wealth. The set  $c(S)$ is the set of all T-period consumption plan.
I could understand why $c(S)$ is compact, but I have no clue why it has to be convex and non-empty.
Following the same question, the professor also mentioned that $W(c)$ the summation of all the T-period utility functions denoted by $W(c)=\sum \limits_{t=1} ^T U(c_t)$ is continuous on $c(S)$. I do not understand why $W(c)$ has to be  continuous on $c(S)$
I think this has something to with Berge's Maximum theorem, but I am unable to link it properly. Any help will be very appreciated.
Assuming the following:

$ S &gt; 0 $, $T &gt; 0 $,
$ c(S) = \left\{c\in\Bbb{R}^T_{+} : \sum_{t=1}^T{c_t} = S \right\} $, and
$ U(\cdot) $ continuous.

Let's start with convexity. To demonstrate convexity, we need to show that for any two points $c^1$ and $c^2$ in $c(S)$, any linear combination of the two is also an element of $c(S)$.
Let $\lambda \in [0,1]$, $c^1 \in c(S)$, $c^2 \in c(S)$. Then, we have
$$ \sum_{t=1}^T{\lambda c^1_t + (1-\lambda) c^2_t} $$
$$ = \sum_{t=1}^T{\lambda c^1_t} + \sum_{t=1}^T{(1-\lambda) c^2_t} $$
$$ = \lambda \sum_{t=1}^T{c^1_t} + (1-\lambda) \sum_{t=1}^T{c^2_t} $$
$$ = \lambda S + (1-\lambda)S $$
$$ = S $$
$$ \Rightarrow \lambda c^1 + (1-\lambda) c^2 \in c(S) $$
Since any linear combination of any two points in c(S) is itself in c(S), we have c(S) convex.
For non-emptiness, we just need to show that a single point exists in c(S):
$$ \sum_{t=1}^T{\frac{S}{T}} = S $$
$$ \Rightarrow \left\{ \frac{S}{T} \right\}^T \in c(S) $$
If $U(\cdot)$ is continuous on $c(S)$, then continuity of $W(c)$ follows from the triangle inequality (see e.g., here or here). If we don't know more about $U(\cdot)$, then there's not much we can say about $W(c)$. Berge's Theorem of the Maximum has to do with the continuity of an optimized function, not the sum of functions.