James's theorem asserts the following: if every continuous linear functional on $E$ attains its norm then $E$ is reflexive. This means that on every non-reflexive space we can find a codimension 1 subspace for which the quotient norm needs the infimum in place of a minimum.
In the comments there were already a few links to threads discussing explicit examples, let me list them for the sake of completeness:
You asked for a finite-dimensional example. This is impossible: for if $F$ is a finite-dimensional subspace of $E$ then we can use compactness of closed and bounded sets in $F$ to ascertain that the infimum is in fact a minimum.
Let $x \in E$ be arbitrary. Choose a sequence $y_{n} \in F$ such that $\lVert x - y_n\rVert \to \lVert x\rVert_{E/F}$. By the reverse triangle inequality $\lVert y_n - x\rVert \geq \lvert \lVert y_n\rVert - \lVert x\rVert\rvert$ we see that $\lVert y_n\rVert \leq 3\lVert x\rVert$ for large enough $n$, so the sequence $(y_n)$ is contained in a bounded subset of $F$. Pass to a convergent subsequence $y_{n_i} \to f \in F$ of $(y_n)$ and observe that $$\lVert x\rVert_{E/F} = \lim \lVert x - y_{n_i}\rVert = \lVert x - f\rVert.$$
With a little more work and refining the above idea, one can prove that a sufficient condition is reflexivity of the subspace $F$.
To see this, recall the following facts:
A norm-closed convex subset $C$ of a Banach space is weakly closed.
By the Hahn-Banach separation theorem we can write $C$ as the intersection of closed half-spaces defined by linear functionals, and these half-spaces are weakly closed since the weak topology is the initial topology induced by the linear functionals.
The closed unit ball of a reflexive Banach space is weakly compact (and hence every closed ball is weakly compact).
Since $E$ is reflexive, it is the dual space of $E^{\ast}$ and the weak and weak*-topologies on $E$ coincide. Since the closed unit ball in a dual space is weak*-compact by Alaoglu's theorem, it is therefore weakly compact.
Combining 1. and 2. we see that every closed and bounded convex set in a reflexive space is weakly compact. [One can also prove that a closed and bounded convex set in a reflexive space is weakly sequentially compact but we won't need this here].
The norm on a Banach space is weakly lower semicontinuous: if a net $x_i$ converges weakly to $x$ then $\lVert x \rVert \leq \liminf_i \lVert x_i\rVert$.
By Hahn-Banach there is a norm 1-functional $\varphi \in E^\ast$ such that $\varphi(x) = \lVert x\rVert$. But then $\lVert x\rVert = \lim_{i} \lvert \varphi(x_i)\rvert \leq \liminf_i \lVert \varphi\rVert\lVert x_i\rVert = \liminf_i \lVert x_i\rVert$.
The restriction of weak topology on $E$ to a closed subspace $F$ of $E$ is the same as the weak topology of $F$ as a Banach space.
This follows from Hahn-Banach.
We are finally ready to prove the announced result:
Let $F$ be a reflexive subspace of the Banach space $E$. Let $x \in E$ be arbitrary. Then there is $f \in F$ such that $\lVert x\rVert_{E/F} = \lVert x - f\rVert = \inf_{y \in F} \lVert x - y\rVert$.
Choose a sequence $y_{n} \in F$ such that $\lVert x - y_n\rVert \to \lVert x\rVert_{E/F}$. Again the reverse triangle inequality shows that $\lVert y_n\rVert \leq 3\lVert x\rVert$ for large enough $n$, so the sequence $(y_n)$ is contained in a weakly compact convex subset of $F$. Pass to a weakly convergent subnet $y_{n_i} \to f \in F$ of $(y_n)$ and observe that $$\lVert x\rVert_{E/F} \leq \lVert x -f\rVert \leq \liminf \lVert x - y_{n_i}\rVert = \liminf \lVert x - y_n\rVert = \lVert x\rVert_{E/F}.$$
So, you want to show that $\operatorname{Lip}(\alpha)$ is a Banach space for the norm $\|\cdot\|_1$.
Let $C[0,1]$ denote the Banach space of continuous complex-valued functions on $[0,1]$, equipped with the sup-norm $\|\cdot\|_\infty$. Then, in particular, for any $f \in \operatorname{Lip}(\alpha)$, $\|f\|_\infty \leq \|f\|_1$.
Now, let $\{f_n\}_{n=1}^\infty$ be a Cauchy sequence in $\operatorname{Lip}(\alpha)$. By the observation above, $\{f_n\}$ is also a Cauchy sequence in $C[0,1]$, and thus converges to some $f \in C[0,1]$ in the sup-norm; it suffices to show that $f \in \operatorname{Lip}(\alpha)$ and that $f_n \to f$ in the norm $\|\cdot\|_1$.
In order to show that $f \in \operatorname{Lip}(\alpha)$, observe that
$$
M_f := \sup_{\left|s-t\right| \neq 0} \frac{\left|f(s)-f(t)\right|}{\left|s-t\right|^\alpha} = \sup_{\epsilon > 0} \sup_{\left|s-t\right|=\epsilon} \frac{\left|f(s)-f(t)\right|}{\left|s-t\right|^\alpha} = \sup_{\epsilon > 0} \epsilon^{-\alpha} \sup_{\left|s-t\right|=\epsilon} \left|f(s)-f(t)\right|,
$$
so that if you can find an upper bound, independent of $\epsilon$, for
$$
M_f^\epsilon := \epsilon^{-\alpha} \sup_{\left|s-t\right|=\epsilon} \left|f(s)-f(t)\right|,
$$
you're in business. Roughly, for any given $\epsilon > 0$, you'll want to find a suitable $f_n$ to make applying the triangle inequality for absolute values work out; it will also help to remember that $\{M_{f_n}\}$ is bounded from above, since $\{\|f_n\|_1\}$ is bounded from above (as $\{f_n\}$ is Cauchy in the norm $\|\cdot\|_1$) and $M_{f_n} \leq \|f_n\|_1$ for each $n$.
Once you do know that $f \in \operatorname{Lip}(\alpha)$, you can probably use similar tricks to get that $f_n \to f$ in the norm $\|\cdot\|_1$.
Best Answer
Let $\{ f_n \}_n$ be a Cauchy sequence in Lip$(\alpha)$ with respect to the norm \begin{align*} \| f \|_2 &:= |f(a)| + M_f \\ &:= |f(a)| + \sup_{t \neq s} \frac{|f(s) - f(t)|}{|s -t|^\alpha}. \end{align*} We wish to show that there exists a function $f \in $Lip$(\alpha)$ such that $\lim_n \| f_n - f \|_2 = 0$. We first note that the sequence of real numbers $\{ f_n(a) \}_n$ is a convergent sequence. Indeed, this follows immediately from the estimate \begin{align*} |f_n(a) - f_m(a)| \leq |f_n(a) - f_m(a)| + M_{f_n - f_m} = \| f_n - f_m \|_2, \end{align*} the assumption that $\{ f_n \}_n$ is a Cauchy sequence in Lip$(\alpha)$, and the fact that $\mathbb R$ is complete. We now claim that $\{ f_n(x) \}_n$ is a convergent sequence in $\mathbb R$ for every $x \in [a,b]$. Indeed, this follows for $x = a$ by what we have already done, while for $x \neq a$, we have by the triangle inequality \begin{align*} |f_n(x) - f_m(x)| &\leq |(f_n(x) - f_m(x)) - (f_n(a) - f_m(a))| + |f_n(a) - f_m(a)| \\ &\leq |x -a|^\alpha M_{f_n - f_m} + |f_n(a) - f_m(a)| \\ &\leq (|x - a|^\alpha + 1) \| f_n - f_m \|_2. \end{align*} This, the Cauchy assumption on $\{ f_n \}_n$ in Lip$(\alpha)$, and the completeness of $\mathbb R$ yield the desired claim. Since the sequence $\{ f_n(x) \}_n$ converges for every $x \in [a,b]$, we may define a function $f:[a,b] \rightarrow \mathbb R$ via $$ f(x) := \lim_{n \rightarrow \infty} f_n(x). $$ We now claim that $f \in$Lip$(\alpha)$. To see this, observe that since $\{ f_n \}_n$ is Cauchy in Lip$(\alpha)$, we have that this sequence is bounded in Lip$(\alpha)$, i.e. $\exists R > 0$ such that $$ \| f_n \|_2 \leq R, \quad \forall n \geq 1. $$ By our definition of $f$, we thus have for $s \neq t$ \begin{align*} \frac{|f(s) - f(t)|}{|s-t|^\alpha} &= \lim_{n \rightarrow \infty} \frac{|f_n(s) - f_n(t)|}{|s-t|^\alpha} \\ &\leq \sup_{n \geq 1} M_{f_n} \\ &\leq \sup_{n \geq 1} \| f_n \|_2 \leq R. \end{align*} Thus, $M_f < \infty$, so that $f \in$ Lip$(\alpha)$. Finally, we show that $\lim_n \| f_n - f \|_2 = 0$. Let $\epsilon > 0$. Since $\{ f_n \}_n$ is Cauchy in Lip$(\alpha)$, there exists $N \geq 1$ such that $$ |f_n(a) - f_m(a)| + M_{f_n - f_m} < \epsilon, \quad n,m \geq N. $$ Let $s \neq t$. Then the previous implies that $$ |f_n(a) - f_m(a)| + \frac{|(f_n(s)- f_m(s)) - (f_n(t)-f_m(t))|}{|s-t|^\alpha} < \epsilon, \quad n,m \geq N. $$ Letting $m \rightarrow \infty$ in the previous inequality implies that $$ |f_n(a) - f(a)| + \frac{|(f_n(s) - f(s)) - (f_n(t) - f(t))|}{|s-t|^\alpha} \leq \epsilon, \quad n \geq N. $$ Taking the supremum over all $s \neq t$ in the previous expression yields $\| f_n - f \|_2 \leq \epsilon$ for all $n \geq N$. This completes the proof.