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For a positive answer to Q1, I came to the same assumptions stated by Bill Johnson in his answer, so I'll adopt his notations.

Let $X$ be a dense, countably generated linear subspace of the separable Banach space $Y$ with unit ball $B_Y.$ Let $T\in L(Y)$ and $\epsilon>0$. Write $X$ as increasing union of a sequence $0=X_0\subset X_1\subset \dots$ of finite dimensional subspaces, with linear projectors $P_n:Y\to X_n$ (in particular, $P_0=0.$)

I think we can choose the projectors $P_n$ (depending on $T$) in such a way that, for every $k$ we have $(I-P_n)T_{|X_k}\to 0$ in the operator norm, as $n\to\infty$. As a consequence, there exists a natural number $n_k$ such that $$\| (I-P_{n_k})\, T\, (P_k-P_{k-1})\|\leq \| (I-P_{n_k})\, T_{|X_k}\|\, \|P_k-P_{k-1}\| \leq \epsilon\ 2^{-k}.$$ The sum $$\sum_{k=1}^\infty\ P_{n_k}\, T\, (P_k-P_{k-1})$$ is punctually finite on $X$, therefore it defines a linear map $T_{\epsilon}:X\to X$ (indeed, it takes $X_k$ into $X_{n_k}$ for every $k$). On the subspace $X$, the operator $T$ also writes in the form $$\sum_{k=1}^\infty\ T\, (P_k-P_{k-1})$$ and one has, on the subspace $X$ $$T-T_{\epsilon}=\sum_{k=1}^\infty\ (I-P_{n_k})\, T\, (P_k-P_{k-1}).$$

By the choice of the sequence $n_k$ the latter series is normally convergent to an operator of norm less than $\epsilon$. Therefore $T_{\epsilon}$ extends to a bounded operator on $Y$ with a distance less than or equal to $\epsilon$ from $T$ such that $T_{\epsilon}(X)\subset\,X.$

The claim should be proved as suggested below by Bill Johnson. Also, a suitable lemma for proving the claim could be stated as follows:

Given the subspaces $\{X_n\}_n$ as above and a countable subset $A\subset Y,$ there are linear projectors $P_n:X\to X_n$ such that $P_na\to a$ as $n\to\infty,$ for all $a\in A.$

Applying this to $A$ equals to the image of a Hamel basis of $X$ via $T$, one has $\|(I-P_n)T_{|X_k}\|=o(1)$ as $n\to\infty$ as we wanted.

Rmk. It seems to me that the statement gains something in generality and semplicity if one considers a different Banach space as codomain: if $T:F\to F'$ is a bounded linear operator; $D\subset F$ is a countable subset; $D'\subset F'$ is dense subset linear subspace; then $T$ can be approximated in operator norm by operators that map $D$ into $D'$ -hence of course $\mathrm{span}(D)$ into $\mathrm{span}(D')$. D'$. This way one sees where the assumptions are needed: countability is only relevant for$D$, density is only relevant for$D'$, and linearity is not really, for both. D'$.

For a positive answer to Q1, I came to the same assumptions stated by Bill Johnson in his answer, so I'll adopt his notations.

Let $X$ be a dense, countably generated linear subspace of the separable Banach space $Y$ with unit ball $B_Y.$ Let $T\in L(Y)$ and $\epsilon>0$. Write $X$ as increasing union of a sequence $0=X_0\subset X_1\subset \dots$ of finite dimensional subspaces, with linear projectors $P_n:Y\to X_n$ (in particular, $P_0=0.$)

I think we can choose the projectors $P_n$ (depending on $T$) in such a way that, for every $k$ we have $(I-P_n)T_{|X_k}\to 0$ in the operator norm, as $n\to\infty$. As a consequence, there exists a natural number $n_k$ such that $$\| (I-P_{n_k})\, T\, (P_k-P_{k-1})\|\leq \| (I-P_{n_k})\, T_{|X_k}\|\, \|P_k-P_{k-1}\| \leq \epsilon\ 2^{-k}.$$ The sum $$\sum_{k=1}^\infty\ P_{n_k}\, T\, (P_k-P_{k-1})$$ is punctually finite on $X$, therefore it defines a linear map $T_{\epsilon}:X\to X$ (indeed, it takes $X_k$ into $X_{n_k}$ for every $k$). On the subspace $X$, the operator $T$ also writes in the form $$\sum_{k=1}^\infty\ T\, (P_k-P_{k-1})$$ and one has, on the subspace $X$ $$T-T_{\epsilon}=\sum_{k=1}^\infty\ (I-P_{n_k})\, T\, (P_k-P_{k-1}).$$

By the choice of the sequence $n_k$ the latter series is normally convergent to an operator of norm less than $\epsilon$. Therefore $T_{\epsilon}$ extends to a bounded operator on $Y$ with a distance less than or equal to $\epsilon$ from $T$ such that $T_{\epsilon}(X)\subset\,X.$

The claim should be proved as suggested below by Bill Johnson. Also, a suitable lemma for proving the claim could be stated as follows:

Given the subspaces $\{X_n\}_n$ as above and a countable subset $A\subset Y,$ there are linear projectors $P_n:X\to X_n$ such that $P_na\to a$ as $n\to\infty,$ for all $a\in A.$

Applying this to the $A$ equals to the image of a Hamel basis of $X$ via $T$, one has $\|(I-P_n)T_{|X_k}\|=o(1)$ as $n\to\infty$ as we wanted.

Rmk. It seems to me that the statement gains something in generality and semplicity if one considers a different Banach space as codomain: if $T:F\to F'$ is a bounded linear operator; $D\subset F$ is a countable subset; $D'\subset F'$ is dense subset; then $T$ can be approximated in operator norm by operators that map $D$ into $D'$ -hence of course $\mathrm{span}(D)$ into $\mathrm{span}(D')$. This way one sees where the assumptions are needed: countability is only relevant for $D$, density is only relevant for $D'$, and linearity is not really, for both.

For a positive answer to Q1, I came to the same assumptions stated by Bill Johnson in his answer, so I'll adopt his notations.

Let $X$ be a dense, countably generated linear subspace of the separable Banach space $Y$ with unit ball $B_Y.$ Let $T\in L(Y)$ and $\epsilon>0$. Write $X$ as increasing union of a sequence $0=X_0\subset X_1\subset \dots$ of finite dimensional subspaces, with linear projectors $P_n:Y\to X_n$ (in particular, $P_0=0.$)

I think we can choose the projectors $P_n$ (depending on $T$) in such a way that, for every $k$ we have $(I-P_n)T_{|X_k}\to 0$ in the operator norm, as $n\to\infty$. As a consequence, there exists a natural number $n_k$ such that $$\| (I-P_{n_k})\, T\, (P_k-P_{k-1})\|\leq \| (I-P_{n_k})\, T_{|X_k}\|\, \|P_k-P_{k-1}\| \leq \epsilon\ 2^{-k}.$$ The sum $$\sum_{k=1}^\infty\ P_{n_k}\, T\, (P_k-P_{k-1})$$ is punctually finite on $X$, therefore it defines a linear map $T_{\epsilon}:X\to X$ (indeed, it takes $X_k$ into $X_{n_k}$ for every $k$). On the subspace $X$, the operator $T$ also writes in the form $$\sum_{k=1}^\infty\ T\, (P_k-P_{k-1})$$ and one has, on the subspace $X$ $$T-T_{\epsilon}=\sum_{k=1}^\infty\ (I-P_{n_k})\, T\, (P_k-P_{k-1}).$$

By the choice of the sequence $n_k$ the latter series is normally convergent to an operator of norm less than $\epsilon$. Therefore $T_{\epsilon}$ extends to a bounded operator on $Y$ with a distance less than or equal to $\epsilon$ from $T$ such that $T_{\epsilon}(X)\subset\,X.$

The claim should be proved as suggested below by Bill Johnson. Also, a suitable lemma for proving the claim could be stated:

Given the subspaces $\{X_n\}_n$ as above and a countable subset $A\subset Y,$ there are linear projectors $P_n:X\to X_n$ such that $P_na\to a$ as $n\to\infty,$ for all $a\in A.$

Applying this to the $A$ image of a Hamel basis of $X$ one has $\|(I-P_n)T_{|X_k}\|=o(1)$ as $n\to\infty$ as we wanted.