Let $B$ be a Banach space with a Schauder basis (thus $B$ is separable), and let $X \subseteq B$ be a closed subspace of $B$ (thus a separable Banach space itself).
Claim: $X$ need not have any Schauder basis.
Proof attempt: Assume by means of contradiction that the claim is false. Let $\tilde{X}$ be the separable Banach space proven by Per Enflo to not have a Schauder basis. By the Banach-Mazur theorem, $\tilde{X}$ is isometrically isomorphic to some closed subspace $X$ of the separable Banach space $B=C([0,1])$. $B=C([0,1])$ has a Schauder basis, e.g. the Faber-Schauder system.
Therefore by the assumption $X \subseteq B=C([0,1])$ has a Schauder basis. Because $\tilde{X}$ is isometrically isomorphic to $X$, $\tilde{X}$ has a Schauder basis, which is a contradiction. $\square$
Possible problem: If $\{e_i\}$ is a Schauder Basis for $B$, and $\phi: B \to \tilde{B}$ an isometric isomorphism, then perhaps $\{\phi(e_i)\} \subseteq \tilde{B}$ need not be a Schauder basis of $\tilde{B}$? I.e. perhaps the property "has a Schauder basis" need not be preserved under isometric isomorphism? Why or why not?
Community wiki: (attempted) proof that isometric isomorphism preserves Schauder basis property.
Let $(V, || \cdot||_V), (W, ||\cdot||_W)$ be Banach spaces over the field $F$ for which there exists an isometric isomorphism $\phi: V \to W$.
Assume that $V$ has a Schauder basis (this implies that $V$ is separable but that's not important to the proof), i.e. there is a sequence $\{\mathbf{b}_N\} \subseteq V$ such that for arbitrary $\mathbf{v} \in V$ there exists a unique (so a well-defined function of $\mathbf{v}$) sequence of scalars $\{\alpha_N\} \subseteq F$ such that $\lim\limits_{N \to \infty} \left|\left|\mathbf{v} - \sum\limits_{n=1}^N \alpha_n \mathbf{b}_n \right|\right|_V = 0$.
Claim: Under these assumptions, the sequence of elements $\{\phi(\mathbf{b}_N)\} \subseteq W$ is a Schauder basis of $W$.
(Ostensible) proof of claim:
Let $ \mathbf{w} \in W$ be arbitrary. Then $\mathbf{w} = \phi(\mathbf{v})$ for some (unique) $\mathbf{v} \in V$, namely $\mathbf{v} := \phi^{-1}(\mathbf{w})$. Because $V$ has the Schauder basis $\{\mathbf{b}_N \}$, there exist a sequence of scalars $\{\alpha_N\} \subseteq F$ such that $\lim\limits_{N \to \infty} c_N = 0$ where we have defined the real sequence $\{ c_N \} \subseteq \mathbb{R}$ as $c_N := \left|\left|\mathbf{v} - \sum\limits_{n=1}^N \alpha_n \mathbf{b}_n \right|\right|_V$ for every $N$.
Because $\phi$ is an isometric isomorphism, it is in particular linear, and therefore for every $N$ we have $$\phi \left(\mathbf{v} - \sum\limits_{n=1}^N \alpha_n \mathbf{b}_n \right) = \phi(\mathbf{v}) - \sum_{n=1}^N \alpha_n \phi(\mathbf{b}_n) = \mathbf{w} - \sum_{n=1}^N \alpha_n \phi(\mathbf{b}_n) \,.$$ Because $\phi$ is an isometric isomorphism, it is in particular an isometry, and so for every $N$ we have $$\left|\left|\mathbf{w} - \sum\limits_{n=1}^N \alpha_n \phi(\mathbf{b}_n) \right|\right|_W = \left|\left|\phi\left(\mathbf{v} - \sum\limits_{n=1}^N \alpha_n \mathbf{b}_n \right) \right|\right|_W = \left|\left|\mathbf{v} - \sum\limits_{n=1}^N \alpha_n \mathbf{b}_n \right|\right|_V = c_N \,.$$ Therefore, given an arbitrary $\mathbf{w} \in W$, we have found a sequence of scalars $\{ \alpha_n \} \subseteq F$ such that $$\lim\limits_{N \to \infty} \left|\left|\mathbf{w} - \sum\limits_{n=1}^N \alpha_n \phi(\mathbf{b}_n) \right|\right|_W = \lim\limits_{N \to \infty} c_N = 0\,. $$ Therefore we will be done if we can show that the sequence $\{\alpha_n \}$ is unique.
So let's say there exists a sequence of scalars $\{ \beta_N \} \subseteq F$ such that $\lim\limits_{N \to \infty}\left|\left|\mathbf{w} - \sum\limits_{n=1}^N \beta_n \phi(\mathbf{b}_n) \right|\right|_W = 0$. Define the sequence of real scalars $\{d_N\} \subseteq \mathbb{R}$ with $\lim\limits_{N \to \infty} d_N = 0$ as $d_N := \left|\left|\mathbf{w} - \sum\limits_{n=1}^N \beta_n \phi(\mathbf{b}_n) \right|\right|_W$ for every $N$.
Based on what we proved above (but applied to the isometric isomorphism $\phi^{-1} : W \to V$ rather than to the isometric isomorphism $\phi: V \to W$) we know that for every $N$ we have $$\left|\left|\mathbf{v} - \sum\limits_{n=1}^N \beta_n \mathbf{b}_n \right|\right|_V = \left|\left| \phi^{-1}(\mathbf{w}) - \sum\limits_{n=1}^N \beta_n \phi^{-1}(\phi(\mathbf{b}_n)) \right|\right|_V = \left|\left|\phi^{-1}\left(\mathbf{w} - \sum\limits_{n=1}^N \beta_n \phi(\mathbf{b}_n) \right) \right|\right|_V = \left|\left|\mathbf{w} - \sum\limits_{n=1}^N \beta_n \phi( \mathbf{b}_n) \right|\right|_W = d_N \,.$$ In particular, we have shown that $\lim\limits_{N \to \infty}\left|\left|\mathbf{v} - \sum\limits_{n=1}^N \beta_n \mathbf{b}_n \right|\right|_V = \lim\limits_{N \to \infty} d_N = 0$. Because of the uniqueness property for the coefficients associated with the Schauder basis $\{\mathbf{b}_n\}$ of $V$, and because $\lim\limits_{N \to \infty} \left|\left|\mathbf{v} - \sum\limits_{n=1}^N \alpha_n \mathbf{b}_n \right|\right|_V = 0$, it follows that the two sequences of scalars $\{\beta_N\} \subseteq F$ and $\{\alpha_N \} \subseteq F$ are equal, $\{\beta_N \} = \{\alpha_N\}$. Therefore we have shown that $\{ \alpha_N \} \subseteq F$ is the only sequence of scalars such that $\lim\limits_{N \to \infty} \left|\left|\mathbf{w} - \sum\limits_{n=1}^N \alpha_n \phi(\mathbf{b}_n) \right|\right|_W = 0$, and therefore verifying the uniqueness property for the scalar coefficients associated with the purported Schauder basis $\{ \phi(\mathbf{b}_N) \} \subseteq W$, thus verifying that $\{ \phi(\mathbf{b}_N) \}$ actually is a genuine Schauder basis of $W$.