Your first question is silly as stated. Consider $0\oplus I$ on $\ell_2 \oplus \ell_\infty$.

As for the second question, Google "total non norming subspaces" to find a wealth of counterexamples.

Added 11/29/16: Oh, so Pietsch does not claim that the subspace $M$ is separable. For your first question to make sense you should assume that $Y$ is separable. Even so, the answer is negative: Consider $0\oplus I$ on $X:= \ell_2\oplus Z$, where $Z$ is a separable space that fails the approximation property (AP). Suppose that $M$ is a subspace of $X$ that has the AP and $\|{Q_M}_{|Z}\| < \epsilon$ for some small enough $\epsilon$ to make the claim below true. Let $P_Z$ be the natural projection from $X$ onto $Z$.

Claim: $P_Z M = Z$.

Assume the claim. Let $W:= \ell_2 \cap M$. Since $W$ is necessarily complemented in $X$, you can write $M$ as the direct sum of $W$ and some closed subspace $M_0$ of $M$, and clearly $P_Z$ is injective on $M_0$ and $P_Z M_0 = Z$. Thus $M_0$ is isomorphic to $Z$ and hence $M_0$, whence also $M$, must fail the approximation property.

Proof of claim. There should be a book reference for this, but I don’t know one. The argument I give is basically the proof of what is often called the ``little open mapping theorem”. By the hypothesis that $\|{Q_M}_{|Z}\| < \epsilon$, for each non zero $z\in Z$ there is $x_z \in M$ s.t. $\| z- x_z\| < \epsilon \|z\|$, and necessarily $\|x_z\| < (1+\epsilon)\|z\|$. Since $\|P_Z\| = 1$, we also have $\|z - P_z x_z\| < \epsilon$. Repeat this, replacing $z$ by $z - P_Z x_z \in Z$ and iterate in the obvious way. This successive approximation argument produces a sequence $(x_n)$ in $M$ s.t. $\|x_n\| < \epsilon^{n-1} (1+\epsilon) \|z\|$ and $z = \sum P_Z x_n$. So you just need $\epsilon < 1$.