Conditions when a subset of compact set is closed. We know that a closed subset of a compact set is compact. But is there any conditions under which a subset of a compact set is closed? 
Actually, I am stumbled upon the topic of Vector-Valued Integration in the article On the Hahn-Banach Theorem by S. Kesavan. It says the following:
Let $\varphi:[0,1]\rightarrow V$ be a continuous mapping into a real Banach space $V$.
Let $H=\mathrm{conv}(\varphi([0,1]))$, i.e., the convex hull, the smallest convex set containing $\varphi([0,1])$.
Then, $\overline{H}=\overline{\mathrm{conv}(\varphi([0,1]))}$.
Then,


*

*$\overline{H}$ is compact by completeness of $V$. 
Like how? I couldn't understand it.
And, again it goes on like this:
Let $L$ be an arbitrary finite collection of continuous linear functionals on $V$. Define
$$E_L=\left\lbrace y\in \overline{H}:f(y)=\int_{0}^{1}f(\varphi(t))\mathrm{d}t\;\mathrm{for\;all}\; f\in L\right\rbrace .$$
And then, it says:
2. It is immediate to see that $E_L$ is a closed set.
How is it possible that $E_L$ is a closed set?
 A: For the second question, define, for $f \in V^{\ast}$, the continuous function $h_f : V \rightarrow \mathbb{R}$ by $h_f(v) = f(v) - \int_{0}^{1}{f(\varphi(t))dt}$ and note that
$$
E_L = \overline{H} \cap \bigcap_{f \in L}{ h_f^{-1}(0)}
$$
is an intersection of closed sets.
For the first question, note that the set $\varphi([0,1])$ is compact  because $[0,1]$ is compact and $\varphi$ is assumed to be continuous. In a Banach space the closure of the convex hull of any compact subset of is compact. To show this, let $K \subset V$ be compact. Then $K$ is totally bounded. It suffices to show that $\text{conv}(K)$ is totally bounded (the precompact subsets of a complete metric space are precisely the totally bounded subsets). Let $\varepsilon >0$. Since $K$ is totally bounded we find a finite set $S \subset K$ so that $K \subset S+ B_{\varepsilon/2}(0)$. Let $C:= \text{conv}{(S)}$. This is a compact convex set (because it can be written as the image of compact set under a continuous map). Since $K \subset C+B_{\varepsilon/2}(0)$ and the latter set is convex, we get $\text{conv}(K) \subset C+B_{\varepsilon/2}(0)$. As $C$ is compact, there is another finite set $F \subset C$ so that $C \subset F+B_{\varepsilon/2}(0)$. It follows that 
$$
\text{conv}(K) \subset C+B_{\varepsilon/2}(0) \subset F + B_{\varepsilon/2}(0)+ B_{\varepsilon/2}(0) \subset F + B_{\varepsilon}(0)
$$
is covered by finitely balls of radius $\varepsilon$, as desired. 
