Existence of an open convex set Let $T$ be a normed vector space, $K\subseteq T$ compact and convex and $O\subseteq K$ convex and open in $K$ (i.e. open w.r.t. the subspace topology of $K$ inherited by $T$).
Can we find an open set $O'\subseteq T$ such that $O'  \cap K = O$? Can we also find such an $O'$ which is convex?
Edit: The convexity of $K$ is indeed necessary. There exists a counterexample in case of $K$ not being convex.
 A: A convex $O'$ need not exist: a counterexample is given by setting $K=[-1,1]\times[0,2]\subseteq\mathbb{R}^2$ and $O=\{(x,y)\in K;y>x^3\}$. Indeed, any open $O'$ with $O'\cap K=O$ would contain some nhood of $(-\frac{1}{2},0)$, so it would contain some point $q=(-\frac{1}{2},-\varepsilon)$ with $\varepsilon>0$. Note that the tangent line from $q$ to the function $f:[0,\infty)\to\mathbb{R};f(x)=x^3$ passes below the point $(0,0)$. This easily implies that if $O'$ was convex, it would contain $(0,0)$, contradicting $O'\cap K=O$.
However, the conjecture is true for lots of convex sets $K$ (for example strictly convex sets, which satisfy the lemma below). I will assume that the normed space is $\mathbb{R}^n$, with $n\geq2$ with norm $||\cdot,\cdot||_2$, since the norm is irrelevant for the problem. First I prove the result for $\overline{\mathbb{B}^n}$, where it's easier to visualize, and then I generalize below.
Proposition: For each convex set $O\subseteq\overline{\mathbb{B}^n}$ open in $\overline{\mathbb{B}^n}$ we can find a convex open set $O'$ such that $O=O'\cap\overline{\mathbb{B}^n}$.
Proof: For each $v\in\mathbb{S}^{n-1}$ let $X_v:=\{x\in\mathbb{R}^n;\langle x,v\rangle<1\}$. Let $B=\mathbb{S}^{n-1}\setminus O$, and let $A$ be the interior of $\bigcap_{v\in B}X_v$. Clearly $A$ is open and convex, and $O\subseteq A$. We will define $O'=O\cup(A\setminus\overline{\mathbb{B}^n})$.
By definition, $O=O'\cap\overline{\mathbb{B}^n}$, and $O'$ is open because $A$ is open, $O\setminus\mathbb{S}^{n-1}$ is open and any $p\in O\cap\mathbb{S}^{n-1}$ is in the interior of $A$, so it is in the interior of $O'$.
So we just need to deduce that $O'$ is convex: let $p,q\in O'$.

*

*If $p,q\in O$ then the segment $[p,q]$ is contained in $O\subseteq O'$.


*If $p\in A\setminus\overline{\mathbb{B}^n},q\in O$, then $[p,q]\subseteq A$. If $[p,q)$ doesn't intersect $\mathbb{S}^{n-1}$, then $[p,q)\subseteq A\setminus\overline{\mathbb{B}^n}\subseteq O'$ so we are done. If not, let $r$ be que unique intersection of $[p,q)$ with $\mathbb{S}^{n-1}$. As $r\in A\cap\mathbb{S}^{n-1}$, $r$ cannot be in $B$, so $r\in O$. Thus $[r,q]\subseteq O\subseteq O'$ and $[p,r)\subseteq A\setminus\overline{\mathbb{B}^n}\subseteq O'$, so as we wanted $[p,q]\subseteq O'$.


*A similar argument in the case $p,q\in A\setminus\overline{\mathbb{B}^n}$ (this time considering the up to $2$ points of intersection of $[p,q]$ with $\mathbb{S}^{n-1}$, which again will be in $O$) allows us to conclude that $O'$ is convex. $\square$
Now let $K$ be a general convex set with nonempty interior. We say an open half space $X$ is tangent to $K$ at $p\in\partial K$ if $K\subseteq\overline{X}$ and $p\in\partial X$. Let $\mathfrak{X}$ be the set of all half spaces tangent to $K$ at points of $\partial K\setminus O$ and let $A$ be the interior of $\bigcap_{X\in\mathfrak{X}}X$. Then again, $A$ is open and convex, but it need not be true that $O\subseteq A$. If indeed $O\subseteq A$, we can still define $O'=O\cup(A\setminus K)$ and the rest of the proof works the same way (when proving convexity, we can define $r$ as the point of $[p,q)\cap\partial K$ closest to $p$).
A sufficient condition for $O\subseteq A$ is the following:
Lemma: Suppose that every hyperplane tangent to $K$ intersects $K$ at just one point. Then $O\subseteq A$.
Proof: Suppose some $p\in O$ is not in $A$. The statement of the lemma implies that $p\in X$ for every $X\in\mathfrak{X}$, so $p\in\bigcap_{X\in\mathfrak{X}}X$. Thus $p$ is in the boundary of $A$, and there is a sequence of half spaces $X_n\in\mathfrak{X}$ tangent to $K$ at some points $p_n\in\partial K\setminus O$, such that $d(p,\partial X_n)\to0$. Taking a subsequence if necessary, we can suppose that the vectors $v_n$ normal to the hyperplanes $\partial X_n$ converge to $v_\infty$, and that the points $p_n$ converge to $p_\infty$ (note that $p_\infty\not\in O$, so $p_\infty\neq p$). So the half space $X$ based at $p_\infty$ with normal vector $v_\infty$ is tangent to $K$, and $p\in\partial X$. Which contradicts the statement of the lemma.$\square$
A: As shown by Saúl RM, the answer is negative. On the other hand, for absolutely convex sets (i.e., in addition to convexity we have $tx\in K$ for all $x\in K$ and $t\in [-1,1]$, if $0\in K$ this is the same as the symmetry $K=-K$) the following weaker statement might be good enough (such tricks are often used in the theory of locally convex spaces):

Proposition. Let $K$ be an absolutely convex symmetric set in a locally convex space $X$ and $U$ a relatively open absolutely convex subset. Then there is an open absolutely convex set $V$ in $X$ with $U\subseteq V\cap K\subseteq 3 U$.

The proof is simple: Since $0\in U$ there is an absolutely convex open $W\subseteq X$ with $W\cap K\subseteq  U$ and we may take $V=U+W$ which is absolutely convex and open in $X$ since so is $W$. The inclusion $U\subseteq V\cap K$ is obvious. For $x=u+w\in V\cap K$ with $u\in U$ and $w\in W$ we have $w=x-u= 2(\frac 12 x + \frac 12 (-u))\in 2K\cap W \subseteq 2(K\cap W) \subseteq 2 U$ so that $x\in U+ 2 U= 3U$ by the convexity of $U$.
