Let $X$ be a Banach space, $K \subset X$ is compact and $$ \text{conv}(K) := \left\{\sum_{i=1}^n t_i x_i : t_i \in [0,1], \sum_{i=1}^n t_i = 1 \text{ and } x_i \in K\right\}$$ be the convex hull of $K$. Show that $\overline{\text{conv}(K)}$ is compact.
Can anyone give some idea on how to start with the proof of the statement?
Note that $$ \text{conv}(K) = \left\lbrace \sum_{i=1}^n t_i x_i : t_i \in [0,1], \sum_{i=1}^n t_i = 1 \text{ and } x_i \in K\right\rbrace $$ Hence, if $K = \{x_1,x_2,\ldots, x_n\}$ is a finite set, then $\text{conv}(K)$ is compact since one can write it as the image of a continuous function $$ f: D \times K_1\times K_2\times \ldots \times K_n\to X $$ where each $K_i = \{x_i\}$ is compact and so is $$ D = \{ (t_1,t_2,\ldots, t_n) : t_i\in [0,1], \sum t_i = 1\} $$
Now suppose $K$ is any compact set, and $\epsilon > 0$, choose finite sets $F \subset K$ and $G\subset \text{conv}(F)$ such that $$ K \subset \bigcup_{x\in F} B(x,\epsilon/2) \text{ and } \text{conv}(F) = \overline{\text{conv}(F)} \subset \bigcup_{y\in G} B(y,\epsilon/2) $$ Now check that $$ \text{conv}(K) \subset \bigcup_{x\in G} B(x,\epsilon) $$ and so $\text{conv}(K)$ is totally bounded, whence $\overline{\text{conv}(K)}$ is compact.