Let $X$ be a normed vector space and $A$ be a subset of $X$. $\operatorname{conv}(A)$ is called the intersection of all convex subsets of $X$ that contain $A$
a) Show that $\operatorname{conv}(A)$ is a convex set
b) Show that
$$\operatorname{conv}(A) = \bigg\{\sum_{i=1}^n\lambda _i x_i : \sum_{i=1}^n \lambda_i =1, \lambda_i\ge 0, x_i\in A, i=1,\cdots,n\bigg\}$$
c) If $A$ is compact then is $\operatorname{conv}(A)$ compact?
d) Show that if $A\subseteq \mathbb{R}^n$ is compact then $\operatorname{conv}(A)$ is compact
a) Take two elements $a$ and $b$ at the intersection of all convex subsets of $X$ that contain $A$. Now take $\lambda a + (1-\lambda)b$. It is contained in every of the subsets of $X$ that contain $A$, therefore it is contained in the intersection. Q.E.D.
b)
[$\Leftarrow$] Suppose by induction that $\sum_{i=1}^n\lambda_i x_i $ belongs to $\operatorname{conv}(A)$. We must prove that $\sum_{i=1}^{k+1}\lambda_i x_i $, with $\sum_{i=1}^{k+1}=1$ also belongs.
$$\sum_{i=1}^{k+1}\lambda_i x_i = \sum_{i=1}^{k}\lambda_i x_i + \lambda_{k+1}x_{k+1} = \sum_{i=1}^{k}\lambda_i x_i + (1-\sum_{i=1}^n\lambda_i)x_{k+1}$$
Now choose $\delta$ such that $\delta\sum_{i=1}^{k}\lambda_i=1$, then
$$\frac{\delta}{\delta} \left(\sum_{i=1}^{k}\lambda_i x_i + (1-\sum_{i=1}^n\lambda_i)x_{k+1}\right)= \left(\frac{1}{\delta}\sum_{i=1}^{k}(\delta\lambda_i) x_i + \frac{1}{\delta}(\delta-1)x_{k+1}\right) =\\ \frac{1}{\delta}x + \left(1-\frac{1}{\delta}\right)x_{k+1}$$
which is a combination of elements of $A$ that sums to $1$
[$\Rightarrow$] Can I just say that $x\in\operatorname{conv}(A)$, then $x = 1x + 0\cdot everything$ therefore it is a combination that sums to $1$ of elements of $A$?
c) A hint is provided: show that $\operatorname{conv}(A\cup B)$ is the image by a continuous function of the compact $\{(\alpha,\beta; \alpha,\beta\ge 0, \alpha + \beta = 1)\}\times A\times B$. I don't understand how to show this.
d) Any hints?