I refer to the answer here by Mariano Suárez-Alvarez.
Question:
$(1)$ Why the set $S = \{ (t_1,t_2,...,t_n) \in \mathbb{R}^n | t_1,t_2,...,t_n \geq 0, t_1 + t_2 + ... + t_n = 1\}$ closed?
$(2)$ Why the map $f:S \rightarrow X$ given by $f(t_1,t_2,...,t_n)=t_1x_1 + t_2x_2 + ... + t_nx_n$ continuous?
For question $(1)$, I understand the set is bounded as in finite dimensional space, all norms are equivalent. So we can take the norm $\| \cdot \|_1$ to show that $\| t \|_1 = \sum_{k=1}^n{t_k} = 1 \leq M$ for all $t \in S$. Hence, $S$ is bounded.
(1) Note that $S$ is the intersection of the closed sets $$ S_1 = \{t \in \mathbf R^n : t_1, t_2, \ldots, t_n \ge 0 \} $$ ($S_1$ is closed, as $S_1 = \bigcap_i\{t_i \ge 0\}$ is the finite intersection of the closed sets $\{t_i \ge 0\}$ which are the continuous preimage of $[0,\infty)$ under the projection $\pi_i \colon \def\R{\mathbf R}\R^n \to \R$, $t \mapsto t_i$) and $$ S_2 = \{t \in \mathbf R^n : \textstyle\sum_i t_i = 1\} $$ ($S_2$ is closed as the continuous preimage of $\{1\}$ under $s \colon t \mapsto \sum_i t_i$).
(2) Note, that addition $+ \colon X^2 \to X$ is continuous due to the triangle inequality and multiplication with scalars due to the positive homogenity of the norm. Hence $f$ is also continuous. To be more explicit, note that $\def\norm#1{\left\|#1\right\|_X}$for $s,t \in \R^n$: \begin{align*} \norm{f(t) - f(s)} &= \norm{\sum_i t_i x_i - \sum_i s_i x_i}\\ &\le \sum_i \def\abs#1{\left|#1\right|}\abs{t_i -s_i}\norm{x_i}\\ &= \sqrt n \left\|{s-t}\right\|_2 \cdot \sum_i \norm{x_i} \end{align*} That is, $f$ is Lipschitz with constant at most $\sqrt n \cdot \sum_i \norm{x_i}$, hence continuous.