Question about relative interior of subsets in $\Bbb R^n$ I would love to get some hint or direction regarding this :

If $S\subseteq T$, when $S,T$ are convex, and ${\rm ri}(S) \cap {\rm ri}(T) \neq  \emptyset$ then ${\rm ri}(S) \subseteq {\rm ri}(T)$.

${\rm aff}(S)$ being the affine subset and ${\rm ri}(S)$ the relative interior.
Thanks!
 A: It is not true.
Take $S=[0,1] \times \{0\}$, $T=[0,1]^2$. Clearly $S \subset T$.
We have $\operatorname{aff}S = \mathbb{R} \times \{0\}$, $\operatorname{aff} T = \mathbb{R}^2$, hence the intersection criterion is satisfied.
However, $\operatorname{ri} S = (0,1) \times \{0\}$, and $\operatorname{ri} T = (0,1)^2$, hence $\operatorname{ri} S \cap \operatorname{ri} T = \emptyset$.
(Note: If $\operatorname{aff}S =\operatorname{aff} T$, then it is true, as then $\operatorname{ri}$ is just the interior with respect to the subspace topology on $\operatorname{aff}S =\operatorname{aff} T$.)
Addendum: Here is an answer to the modified question:
The key here is the following (as an aside, I think this result encapsulates one of the reasons why convexity is such a big deal, that is, behaviour at a single point can be translated, in some sense, to behaviour at many other points):
Lemma: Suppose $C$ is convex, $x \in \operatorname{ri} C$ and $y \in C$. Then $\lambda y +(1-\lambda)x \in \operatorname{ri} C$ for all $\lambda\in[0,1)$.

The picture illustrates why this is true. The proof is also straightforward; for some $\epsilon>0$ we have $B(x,\epsilon) \cap \operatorname{aff} C \subset C$. Let $\lambda \in [0,1)$. Then I claim $B(\lambda y +(1-\lambda)x,(1-\lambda)\epsilon) \cap \operatorname{aff} C \subset C$. Suppose $p \in B(\lambda y +(1-\lambda)x,(1-\lambda)\epsilon) \cap \operatorname{aff} C$. Then we have $p \in \operatorname{aff} C$ and $\|p-(\lambda y +(1-\lambda)x) \| < (1-\lambda)\epsilon$, or, since $1-\lambda >0$, we have $\|\frac{p-\lambda y}{1-\lambda} -x) \| < \epsilon$. Since $\frac{p-\lambda y}{1-\lambda} \in \operatorname{aff} C$, we have $\frac{p-\lambda y}{1-\lambda} \in C$, and since $p = \lambda y + (1-\lambda) \frac{p-\lambda y}{1-\lambda}$, we have $p \in C$ and so we have the desired result.
Now suppose we have some $x \in \operatorname{ri} S \cap \operatorname{ri} T $ and let $y \in \operatorname{ri} S \subset T$. To finish, we need to show that $y \in \operatorname{ri} T$ as this will show that $\operatorname{ri} S \subset 
\operatorname{ri} T$.
Since $y \in \operatorname{ri} S$, then for some $\mu>1$ we have $w=\mu y + (1-\mu)x \in S$ (hence $w \in T$). Since $y = \frac{1}{\mu} w + (1-\frac{1}{\mu}) x$ and $x \in \operatorname{ri} T$, the lemma above shows that $y \in \operatorname{ri} T$, as desired.
A: Probably you no longer need an answer (after 5 years), but still:
It is well-known (proof in many textbooks on convex analysis) that $\mathop{\rm ri}(S)\cap\mathop{\rm ri}(T)\neq\emptyset$ implies $\mathop{\rm ri}(S)\cap\mathop{\rm ri}(T)=\mathop{\rm ri}(S\cap T)$. If $S\subseteq T$, which is the same as $S\cap T=S$, then
$\mathop{\rm ri}(S)\cap\mathop{\rm ri}(T)=\mathop{\rm ri}(S\cap T)=\mathop{\rm ri}(S)$, which implies $\mathop{\rm ri}(S)\subseteq\mathop{\rm ri}(T)$.
