Theorem 2.5-5 from the Introductory functional Analysis by Erwin Kreyszig 
I don't understand why $X_1$ has dimension one? (underlined in the text).
Also, Riesz's 
lemma says that $x_2$ must belong to a set of which $X_1$ is a proper subset of it; how it can be a proper subset if both has the property of their norms being equal (=1)? Thanks.
 A: $X_1$ is the span of the single vector $x_1$ so has basis $\{ x_1 \}$ and hence is one dimensional.
Edit: To answer the second question that you added - $X_1$ is a closed proper subspace of $X$ since $\mbox{dim}(X) = \infty$ so we are able to apply Riesz' lemma to the subspace $X_1$ of $X$ to find a vector $x_2 \not \in X_1$ with the desired distance property. Note that $X_1$ contains exactly one vector of norm $1$ whilst $X$ contains infinitely many.
A: Note that $X_2=span\{x_1,x_2\}=\{ax_1+bx_2|a,b \in \mathbb{F}\}$  
and $X_1=span\{x_1\}=\{cx_1|c \in \mathbb{F}\}$
Now $X_1 \subseteq X_2$ because evey evely element of $X_1$ is has the form $ax_1+0x_2 \in X_2$ where $a,0 \in \mathbb{F}$
Also note that in a normed vector  space $X$ it is possible to find elements $x_1,x_2 \in X$ such that $||x_1||=||x_2||$ but this does not always imply that $x_1=x_2$
Take for instance $\mathbb{R}^3$ with the usual norm $||.||_2$
and $x_1=(1,0,0)$ and $x_2=(0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$ which are also linearly independent.
We have that $||x_1||_2=||x_2||_2=1$ but $x_1 \neq x_2$  and $$A=span\{x_1\} \subsetneq B=span\{x_1,x_2\}$$
