I have to prove two items, but I only got a part. (all sets are subsets of $\mathbb{R}^{n}$).
Prove that
(a) If all $Y \subset \mathbb{R}^{n}$ homeomorphic to $X$ is closed, so $X$ is compact.
(b) Let $C(X)$ be the intersection of all convex subsets $C$ of $\mathbb{R}^{n}$ with $C \supset X$. If $X$ is compact, show that $C(X)$ is compact.
For (a). $X$ is homeomorphic to $X$ by the identity aplication, so $X$ is closed. I don't see how to use the hypothesis for show that $X$ is bounded. Any hint?
For (b). Since $X$ is compact, $X \subset B(0,\delta)$ for some $\delta > 0$. But $B(0,\delta)$ is convex, so $C(X) \subset B(0,\delta) \therefore C(X)$ is bounded. I know that I should to use convexity for show that $C(X)$ is compact, maybe, to use that given $x \in C(X)$, $x$ is convex combination. Any hint?
For (a), provided that there exists at least one $Y\subset \Bbb R^n$ which is homeomorphic to $X:$ For $x=(x_1,...,x_n)\in \Bbb R^n$ let $g(x)=(\arctan x_1,..., \arctan x_n).$ Then $g:\Bbb R^n\to (-\pi/2,\pi/2)^n$ is a homeomorphism.
If $Y\subset \Bbb R^n$ is closed in $\Bbb R^n$ but is not compact then $Y$ is unbounded, so $g(Y)$ is not closed in $\Bbb R^n$, but is homeomorphic to $Y$.
So any $Y\subset \Bbb R^n$ that satisfies the hypothseses of the Q is compact. So if such a $Y$ exists then $X$ is homeomorphic to a compact space $Y,$ so $X$ is compact.
Remark: If $Y\subset \Bbb R^n$ and $g(Y)$ is closed in $\Bbb R^n$ then ( since $g(Y)$ is also bounded ) $g(Y)$ is compact. Then ( since $g^{-1}$ is continuous ) $Y=g^{-1}(g(Y))$ is compact , and hence $Y$ is bounded.