Infinite intersection of $L^p$ space Consider the space
$$L^{2-}(\mathbb{R}):=\bigcap_{1\leq p<2}L^{p}(\mathbb{R})$$
equipped with the norm
$$\Vert f\Vert_{L^{2-}(\mathbb{R})}:=\sup_{1\leq p<2}\Vert f\Vert_{L^{p}(\mathbb{R})}.$$
Is it true that $(L^{2-}(\mathbb{R}),\Vert\cdot\Vert_{L^{2-}(\mathbb{R})})$ is a Banach space?

No, as your norm is not well-defined on the space. Consider the function $f(x) = x^{-1/2} 1_{(0,1)}(x),$ then for $p<2$ we have,
$$ \lVert f \rVert_{L^p(\mathbb R)} = \left(\int_0^1 x^{-p/2} dx\right)^{1/p} = \frac1{(1-p/2)^{1/p}},$$
so $f \in L^p(\mathbb R)$ for each $1 \leq p <2,$ but $\lVert f \rVert_{L^{2-}(\mathbb R)} = \infty.$
Note if you instead consider the space
$$ X = \left\{ f \in \bigcap_{1\leq p < 2} L^p(\mathbb R) : \lVert f \rVert_{L^{2-}(\mathbb R)} < \infty \right\},$$
then this is a Banach space with the given norm and moreover it coincides with $L^1(\mathbb R) \cap L^2(\mathbb R).$
Edit: To elaborate on my comment on the end, we can show that $X = L^1(\mathbb R) \cap L^2(\mathbb R)$ as follows:
First if $f \in X,$ then noting $f^p \rightarrow f^2$ as $p \rightarrow 2$ pointwise, we get by Fatou's lemma,
$$ \int_{\mathbb R} |f|^2 dx \leq \liminf_{p \rightarrow 2^-} \int |f|^p dx \leq \lVert f \rVert_{L^{2-}(\mathbb R)}^2, $$
so $f \in L^2(\mathbb R).$ Conversely if $1<p<2$ and $f \in L^1(\mathbb R) \cap L^2(\mathbb R)$ choosing $\alpha \in (0,1)$ such that $\frac1p = \frac{\alpha}{1} + \frac{1-\alpha}{2}$ we get by Hölder's inequality,
$$ \lVert f \rVert_{L^p(\mathbb R)} \leq \lVert f \rVert_{L^1(\mathbb R)}^{\alpha} \lVert f \rVert_{L^2(\mathbb R)}^{1-\alpha}. $$
So $f \in X.$ So $X = L^1(\mathbb R) \cap L^2(\mathbb R).$
