$Let\ \mathbb{X}$ be a normed space. Its dual space is $\mathbb{X}' := \mathcal{L}(\mathbb{X};\mathbb{K}), $ equipped with the operator norm. Denote $\mathbb{X''}:= \ (\mathbb{X}')'.$ Define the map $\tau : \mathbb{X} \rightarrow \mathbb{X''} $ as follows: for $x \in \mathbb{X}$, $\tau(x)$ is the function from $\mathbb{X'}$ to $\mathbb{K}$ which on input $f \in \mathbb{X}'$ outputs $f(x)$, that is, $\tau(x)(f):=f(x)$.
1. How can I prove this map is well-defined? My understanding is that I should show that for each $x \in \mathbb{X}$, $\tau(x)$ is a linear map $\mathbb{X}' \rightarrow \mathbb{K}$ with a bounded operator norm.
2. Why does the linear map $\tau:\mathbb{X} \rightarrow \mathbb{(X')'}$ have operator norm at most $1$ and how could I show it?
There is no problem with the definition of $\tau$: for each $x\in X$, you get $\tau(x)\in X''$; the only thing to check is that it is bounded. For that, you have, for any $x\in X$ and $f\in X'$, $$ |\tau(x)f|=|f(x)|\leq\|x\|\,\|f\|. $$ Since $f$ was arbitrary, this shows that $\|\tau(x)\|\leq\|x\|$; and, as $x$ is arbitrary, $\|\tau\|\leq1$.
Now fix $x\in X$, with $\|x\|=1$. By Hahn-Banach, $$ \|x\|=\sup\{|f(x)|:\ f\in X',\ \|f\|=1\}. $$ So for each $\varepsilon>0$ there exists $f\in X'$, with $\|f\|=1$ and $|f(x)|>1-\varepsilon$. Then $$ |\tau(x)f|=|f(x)|>1-\varepsilon, $$ which implies $\|\tau(x)\|>1-\varepsilon$. And this, in turn, implies that $\|\tau\|>1-\varepsilon$. We can do this for all $\varepsilon>0$, so $\|\tau\|\geq1$. Thus $\|\tau\|=1$.