Lemma: If $Y$ is a closed proper subspace of $X$, then for every $\varepsilon > 0$, there exists $x \in X$ with $\|x\| = 1$ such that $$ d(x,Y) > 1 - \varepsilon $$
Proof: Since $Y$ is a closed proper subset of $X$, by a corollary of Hanh-Banach theorem, there exists an $f \in X^{\ast}$ with $\|f\| = 1$ and $f|_Y = 0$.
Since $\|f\| = 1$, for every $\varepsilon > 0 $, there exists $x \in X$ with $\|x\| = 1$ $$ |f(x)| > 1 - \varepsilon $$
The section in the rectagle is the part of the proof I'm having hard time to understand.
If $\|x\| = 1$ then we have:
$$ |f(x)| \leq \|f\|\|x\|= 1 \cdot 1=1 $$ and this ensures that $|f(x)|$ is bounded by $1$.
How did we obtain that $|f(x)|$ is arbitrarily close to $1$? Is it just a more formal way to say that $1$ is an upper bound?
Remember that $\|f\|$ is a supremum $$ \|f\|=\sup_{x:\|x\|=1}|f(x)|. $$ Being the supremum implies that there are values in your set arbitrarily close to it.