Let $E$ a normed vector space and $M\subseteq E$ a convex set. Then prove that $M^\circ$ and $\bar M$ are convex.
(Here I used $M^\circ$ to denote the interior set of $M$, and $\bar M$ to denote the closure of $M$.)
I would like a verification for this proof. Thank you in advance.
Lets start with the case $M$ convex implies $M^\circ$ is also convex. Suppose that $M^\circ$ is not convex, then for some $x,y\in M^\circ$ the path defined for convexity as
$$f:[0,1]\to E,\quad t\mapsto (1-t)x+ty$$
is not contained in $M^\circ$ (but it is contained in $M$ because $M$ is convex), that is, exists some $t'$ such that $f(t')\notin M^\circ$, i.e. $f(t')$ is not an interior point of $M$, then for any open ball of $f(t')$ there are points that belong to $M^\complement$.
Because $x$ and $y$ are interior points choose some $\epsilon>0$ such that the open balls $\Bbb B(x,\epsilon)$ and $\Bbb B(y,\epsilon)$ are contained in $M^\circ$.
Then we pick some $p\in(\Bbb B(f(t'),\epsilon)\cap M^\complement)$, i.e. some point that is not contained in $M$ and belong to the open ball of $f(t')$ of radius $\epsilon$.
Now set $v:=f(t')-p$. Now observe that the vectors $x-v$ and $y-v$ belong to the respective balls defined above for $x$ and $y$, i.e.
$$\|v\|=\|f(t')-p\|<\epsilon\implies \|x-v\|< \|x\|+\epsilon\implies x-v\in\Bbb B(x,\epsilon)$$
Then the straight path defined between $x-v$ and $y-v$ contains $p$, what not belong to $M$, i.e.
$$g(t')=(1-t')(x-v)+t'(y-v)=(1-t')x+t'y-(1-t')(f(t')-p)-t'(f(t')-p)=p$$
where $f(t')=(1-t')x+t'y$. Then we have a contradiction with the condition of convexity over $M$. Then $M^\circ$ is convex.$\Box$
For the case of $\bar M$ we repeat the same technique, i.e. if $\bar M$ is not convex then exists some $x,y\in\bar M\setminus M$ such that for some $t'$ we have that $f(t')$ is not contained in $\bar M$.
Then we choose some $\epsilon>0$ such that $\Bbb B(f(t'),\epsilon)\subset \bar M^\complement$, because $f(t')$ is an interior point of $\bar M^\complement$. Now we pick points $x'\in(\Bbb B(x,\epsilon)\cap M)$ and $y'\in(\Bbb B(y,\epsilon)\cap M)$.
In a similar manner as above we can conclude that the point $g(t')$ defined as
$$g(t')=(1-t')x'+t'y'$$ is contained in the open ball $\Bbb B(f(t'),\epsilon)$, i.e.
$$\|(1-t')x'+t'y'\|=\|(1-t')(x+r)+t'(y+s)\|<\|f(t')\|+\epsilon$$
because $\|s\|,\|r\|<\epsilon$. This contradicts the assumption that $M$ is convex, so $\bar M$ is convex.$\Box$
It looks good. However it is a little bit complicated than necessary.
For the interior: Let $x,y\in M^\circ$ with $x\ne y$ and $t\in(0,1)$. Then, the mapping, for $u\in E$, $$ f(u) = (1-t)x + tu $$ is a homeomorphism from $E$ onto $E$, as $t\ne 0$. By convexity of $M$ we also have $f(M) \subseteq M$. Now, let $V\subseteq M$ be an open neighborhood of $y$, then $f(V)$ is an open neighborhood of $f(y)$ and subset of $M$.
For the closure: Let $x,y\in \bar M$ with $x\ne y$ and $t\in(0,1)$. Let $x_\alpha, y_\alpha\in M$ with $(x_\alpha, y_\alpha)\to (x,y)$. Then, $$ M \ni (1-t) x_\alpha + ty_\alpha \to (1-t)x + ty \in \bar M. $$