Prove that a point is always in the interior of a compact Probably a very simple proof but I can't write it properly, although I can imagine it.

Let $F$ a finite-dimensional normed vector space and $A \subset F$.

How to prove that for $a \in A$, we can find a compact $K \subset A$ so that $a \in K^{\circ}$ where $K^{\circ}$ is the interior of $K$ ?
Going to slightly change notation:

note that for an arbitrary Hausdorff space (which all normed spaces are), compact implies locally compact. The "worst" case scenario is just taking the whole space to the compact neighborhood, and so indeed, compact implies locally compact (the converse of course fails, just look at $\mathbb{R}$)

So let $K$ be some some compact set, and let $a\in K$ be an arbitrary element. We know that there is a compact neighborhood $E$ around the point $a$. Note that $E$ is closed, because compact subsets in a Hausdorff space (and hence normed spaces) are closed. So now define the open set/ball $U=int(E)$. We want to show this is non-empty. 

Here is where we used properties of normed spaces. Our space is connected because every finite dimensional vector space is isomorphic to $\mathbb{R}^n$ which is both connected and locally connected. But we want a homeomorphism, not just an isomorphism. So we can just take a compact set $X\in \mathbb{R}^n$ and define the map $\phi:X\rightarrow K$. 

We need this map to be continuous and bijective, because a continuous bijection between a compact set $X$ and a Hausdorff space $K$ is a homeomorphism. This will allow us to treat $K$ as topologically equivalent to $X$, so topological properties true for $X$ will hold for $E$. (Just let this be some arbitrary linear, bijective map between basis vectors. 

A restriction of a continuous map is continuous, so if it holds for $K$ it will hold for $E$. We have now found a homeomorphism, so topological properties such as connectedness are preserved, so we now know $E$ is connected. 

Normed spaces must have more than one point for the norm to make sense, but all metric spaces with more than one point are uncountable and all normed spaces are also metric spaces. So our set $E$ is also uncountable. 

Since $E$ is compact Hausdorff, we can find disjoint open sets $\mathcal{U}, \mathcal{V}$ which separate $a$ from another point $b$. In particular, let $\mathcal{U}=U$ and so we have found that $a$ is an interior point of some compact subset $E$, embedded in a normed space $K$
Going to slightly change notation:

note that for an arbitrary Hausdorff space (which all normed spaces are), compact implies locally compact. The "worst" case scenario is just taking the whole space to the compact neighborhood, and so indeed, compact implies locally compact (the converse of course fails, just look at $\mathbb{R}$)

So let $K$ be some some compact set, and let $a\in K$ be an arbitrary element. We know that there is a compact neighborhood $E$ around the point $a$. Note that $E$ is closed, because compact subsets in a Hausdorff space (and hence normed spaces) are closed. So now define the open set/ball $U=int(E)$. We want to show this is non-empty. 

Here is where we used properties of normed spaces. Our space is connected because every finite dimensional vector space is isomorphic to $\mathbb{R}^n$ which is both connected and locally connected. But we want a homeomorphism, not just an isomorphism. So we can just take a compact set $X\in \mathbb{R}^n$ and define the map $\phi:X\rightarrow K$. 

We need this map to be continuous and bijective, because a continuous bijection between a compact set $X$ and a Hausdorff space $K$ is a homeomorphism. This will allow us to treat $K$ as topologically equivalent to $X$, so topological properties true for $X$ will hold for $E$. (Just let this be some arbitrary linear, bijective map between basis vectors. 

A restriction of a continuous map is continuous, so if it holds for $K$ it will hold for $E$. We have now found a homeomorphism, so topological properties such as connectedness are preserved, so we now know $E$ is connected. 

Normed spaces must have more than one point for the norm to make sense, but all metric spaces with more than one point are uncountable and all normed spaces are also metric spaces. So our set $E$ is also uncountable. 

Since $E$ is compact Hausdorff, we can find disjoint open sets $\mathcal{U}, \mathcal{V}$ which separate $a$ from another point $b$. In particular, let $\mathcal{U}=U$ and so we have found that $a$ is an interior point of some compact subset $E$, embedded in a normed space $K$