= Solution
Write $L(f)=\sup_{x\ne y}|f(x)-f(y)|/d(x,y)$ and $B(f)=\sup_z|f(z)|$. The quantity in the question is the <bounded Lipschitz norm> $\|f\|_{\rm BL}=L(f)+B(f)$. Both terms are nonnegative and absolutely homogeneous, and each satisfies the <triangle inequality>. Therefore their sum does too; finite values are preserved by addition and scalar multiplication, making the domain a <vector space>. If the sum is zero, $B(f)=0$ forces $f=0$. Thus it is a <norm>.
For the sequence bounded by one, both $L(f_i)\leq1$ and $B(f_i)\leq1$. Given real $x$, choose a nearby rational $q$. Then
$$
|f_i(x)-f_j(x)|\leq2|x-q|+|f_i(q)-f_j(q)|.
$$
First make $|x-q|$ small, then use convergence at $q$ to make the last term small. The values at every real $x$ form a <Cauchy sequence>, so $f_i$ converges pointwise to a real function $f$.
To preserve the sum bound rather than merely bound its two terms separately, use every fixed triple $x\ne y,z$:
$$
\frac{|f_i(x)-f_i(y)|}{|x-y|}+|f_i(z)|\leq1.
$$
Pass to the pointwise limit and take the independent suprema over $(x,y)$ and $z$. This gives $\boxed{\|f\|_{\rm BL}\leq1}$, the <pointwise closure of the bounded Lipschitz unit ball>.
To obtain <diagonal compactness for bounded Lipschitz functions>, enumerate the rationals for an arbitrary sequence in this unit ball. Values at the first rational lie in $[-1,1]$, so the <Bolzano-Weierstrass theorem> supplies a convergent subsequence there. Repeatedly extract nested subsequences for successive rationals. The <diagonal subsequence argument> gives one subsequence converging at every rational. Applying the preceding argument proves \b[there is a pointwise convergent subsequence with limit of bounded Lipschitz norm at most one].
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