By definition $X'$ is a space of bounded linear functionals on $X$. More preciesly
$$
X'=\{f\in\mathcal{L}(X,\mathbb{C}):\Vert f\Vert<+\infty\}
$$
where $\mathcal{L}(X,\mathbb{C})$ is a linear space of linear functions from $X$ to $\mathbb{C}$ and
$$
\Vert f\Vert:=\sup\{|f(x)|:x\in X\quad \Vert x\Vert\leq 1\}
$$
In order to prove that $X'$ is complete consider Cauchy sequence $\{f_n:n\in\mathbb{N}\}\subset X'$. Fix $\varepsilon>0$. Since $\{f_n:n\in\mathbb{N}\}$ is a Cauchy sequence there exist $N\in\mathbb{N}$ such that for all $n,m>N$ we have $\Vert f_n-f_m\Vert\leq\varepsilon$. Consider arbitrary $x\in X$, then
$$
|f_n(x)-f_m(x)|=|(f_n-f_m)(x)|\leq\Vert f_n-f_m\Vert\Vert x\Vert\leq\varepsilon \Vert x\Vert
$$
Thus we see that $\{|f_n(x)|:n\in\mathbb{N}\}\subset\mathbb{C}$ is a Cauchy sequence. Since $\mathbb{C}$ is complete, there exist unique $\lim\limits_{n\to\infty}f_n(x)$. Since $x\in X$ is arbitrary we can define function
$$
f(x):=\lim\limits_{n\to\infty}f_n(x)
$$
Our aim is to show that $f\in X'$ and $\lim\limits_{n\to\infty}f_n=f$.
Let $x_1,x_2\in X$, $\alpha_1,\alpha_2\in\mathbb{C}$ then
$$
f(\alpha_1 x_1+ \alpha_2 x_2)=
\lim\limits_{n\to\infty}f_n(\alpha_1 x_1+ \alpha_2 x_2)=
$$
$$
\lim\limits_{n\to\infty}(\alpha_1 f_n(x_1) + \alpha_2 f_n(x_2))=
\alpha_1 \lim\limits_{n\to\infty}f_n(x_1) + \alpha_2 \lim\limits_{n\to\infty}f_n(x_2))=
\alpha_1 f(x_1) + \alpha_2 f(x_2)
$$
So we conclude $f\in\mathcal{L}(X,\mathbb{C})$. Since $\{f_n:n\in\mathbb{N}\}$ is a Cauchy sequence it is bounded in $X'$, i.e. there exist $C>0$ such that $\sup\{\Vert f_n\Vert:n\in\mathbb{N}\}\leq C$. Hence, for all $x\in X$ we have
$$
|f(x)|=
|\lim\limits_{n\to\infty}f_n(x)|=
\lim\limits_{n\to\infty}|f_n(x)|\leq
\limsup\limits_{n\to\infty}\Vert f_n\Vert \Vert x\Vert\leq
\Vert x\Vert\sup\{\Vert f_n\Vert:n\in\mathbb{N}\}\leq
C\Vert x\Vert
$$
Now we see that $\Vert f\Vert\leq C$, but as we proved earlier $f\in\mathcal{L}(X,\mathbb{C})$, so $f\in X'$.
Finally recall that for given $\varepsilon>0$ and $x\in X$ there exist $N\in\mathbb{N}$ such that $n,m>N$ implies
$$
|f_n(x)-f_m(x)|\leq\varepsilon \Vert x\Vert.
$$
Then let's take here a limit when $m\to\infty$. We will get
$$
|f_n(x)-f(x)|\leq\varepsilon \Vert x\Vert.
$$
Since $x\in X$ is arbitrary we proved that for all $\varepsilon>0$ there exist $N\in\mathbb{N}$ such that $n>N$ implies
$$
\Vert f_n-f\Vert=
\sup\{|f_n(x)-f(x)|:x\in X,\quad \Vert x\Vert\leq 1\}\leq
\varepsilon.
$$
This means that $\lim\limits_{n\to\infty} f_n=f$. Since we showed that every Cauchy sequence in $X'$ have a limit, $X'$ is complete.
This proof can be easily generalized up to the following theorem: If $X$ is a normed space and $Y$ is a Banach space, then the linear space of all bounded linear functions from $X$ to $Y$ is complete.
My advice is to place a lot more landmarks like $\mathbb R^n$. Ideally, every area should have at least one point in it, which will serve to prove that the area really belongs there. It will also clarify what the relationships really mean. For example, all manifolds are metrizable, but not uniquely. So if you want "manifolds" to extend outside of "metric spaces", then you should add a landmark like $S^1$ and then, in a list of landmarks below the diagram, explain why it's there:
$S^1$ denotes the circle as a topological space. It is a manifold. It is not homeomorphic to any real vector space, since it is compact. It is metrizable, like all manifolds, but it doesn't come equipped with any particular metric.
Speaking of which, manifolds have a finite dimension, which is a topological invariant. So if a real manifold does have a real vector space structure, then it is a finite-dimensional vector space, and that may make it difficult to draw meaningful distinctions within all the little slices in the manifold box. Again, depending on what you really mean, you might be able to justify those slices, so I'm not going to say that they're wrong. Trying to place landmarks in there will force you to decide what you want them to mean.
Once you go through enough examples, then you can summarize the meanings in a preface to the diagram:
This diagram depicts X. One box is placed entirely inside another box if either Y or (when it makes sense) Z.
Best Answer
Let $(B, || \cdot||)$ be a Banach space. Then $d(x,y):=||x-y||$ is a translation-invariant metric on $B$ and $B$ is complete with respect to this metric.
A basis of neighborhoods of $0$ is given by the sets $B_{\epsilon}:=\{x \in B:||x||< \epsilon\}, $ where $ \epsilon >0$.
Each $B_{\epsilon}$ is convex.
Conclusion: a Banach is a Frechet space.