Give an example of an infinite family of convex sets in $\mathbb{R}^d$ such that any $d + 1$ of them have a common point but all of them do not have. It is enough to find such a family for some $d$.
I am not able to understand the question here. Could someone please say what does this mean and what will be its answer?
I suppose that "having a common point" means they share at least one point.
Here is an answer on $\mathbb{R}$, I will let you do the job to expand the reasonning to arbitrary dimensions but it is not that hard if you think correctly...
Define $I_n = (0, \frac 1 n)$. Each $I_n$ is convex and any intersection of $2$ of them have a common point. But if you make the intersection of the whole family it will be empty.
Indeed, let $A$ be that intersection and suppose $x \in A$. Then $x< \frac 1n$ for all $n$ so $x\le 0$ but every non-positive element is not in $I_1=(0,1)$ for example. Hence a contradiction.