Just to add an answer which explicitly uses the Approximation property.

A Banach space has the approximation property (AP) when, for $x_1,\cdots,x_n \in E$ we can find a net of finite-rank operators $T_i$ such that $T_i(x_k)\rightarrow x_k$ for each $k$.

In our case, $\iota:H\rightarrow H_0$ is a continuous map with dense range. If $H_0$ has the AP then for $x_1,\cdots,x_n \in H_0$ and $\epsilon>0$ we can find $T:H_0\rightarrow H_0$ finite rank with $\| T(x_i) - x_i \| < \epsilon$.

As $T(H_0)$ is finite dimensional and $\iota$ has dense range, we can find a linear map $S: T(H_0) \rightarrow H$ so that that $\|\iota S(x) - x\| \leq \epsilon$ for all $x$ in the unit ball of $T(H_0)$. [Proof: If $M\subseteq H_0$ is finite-dimensional, with linear basis $m_1,\cdots,m_n$, then as all norms are equivalent on $M$, if we can ensure that $\|\iota S(m_i)-m_i\|$ is very small, then $\|\iota S(x)-x\|$ will be small uniformly on the unit ball of $M$. But this follows as $\iota$ has dense range and we can choose each $S(m_i)$ completely freely.]

Then $\| \iota ST(x_i) - x_i\| \leq \| \iota ST(x_i) - T(x_i) \| + \| T(x_i) - x_i\| < \epsilon\|T(x_i)\| + \epsilon$ $< \epsilon^2 + \epsilon\|x_i\| + \epsilon$. So $ST : H_0 \rightarrow H$ approximates the identity in the $H_0$ norm. In this way we obtain a net (if $H_0$ is separable we can use a sequence) as required.

Remark 1: Having the "compact approximation property" doesn't seem to help. By definition, this means we can only choose $T$ to be compact not finite-rank. Then the image of the unit ball of $H_0$ under $T$ is a compact set, but I don't know how to form the equivalent of $S$. That is, how do you (linearly) distort a compact set from $H$ into $H_0$?

Remark 2: For Frechet spaces, the argument should be similar, but working with the countable family of seminorms. But I haven't checked the details.

Edit: So I think my real mistake was in the claim that "if $H_0$ is separable then we can use a sequence". As Bill Johnson implictly points out, you can always find a net $P_\alpha:H_0\rightarrow H$.

Just to correct the argument (though Martins now gives it in more generality)... If $H_0$ has the bounded approximation property, then there is an absolute constant $\lambda>0$ so that for $x_1,\cdots,x_n\in H_0$ there is a finite-rank operator $T:H_0\rightarrow H_0$ with $\|T\|\leq \lambda$ and $\|T(x_i)-x_i\| \leq \epsilon$ for each $i$.

In our case, $\iota:H\rightarrow H_0$ is a continuous map with dense range. For $x_1,\cdots,x_n \in H_0$ and $\epsilon>0$ we find a finite-rank $T$ with $\|T\|\leq\lambda$ and $\|T(x_i)-x_i\| \leq \epsilon$ for each $i$.

As $T(H_0)$ is finite dimensional and $\iota$ has dense range, we can find a linear map $S: T(H_0) \rightarrow H$ so that that $\|\iota S(x) - x\| \leq \epsilon$ for all $x$ in the unit ball of $T(H_0)$. [Proof: If $M\subseteq H_0$ is finite-dimensional, with linear basis $m_1,\cdots,m_n$, then as all norms are equivalent on $M$, if we can ensure that $\|\iota S(m_i)-m_i\|$ is very small, then $\|\iota S(x)-x\|$ will be small uniformly on the unit ball of $M$. But this follows as $\iota$ has dense range and we can choose each $S(m_i)$ completely freely.]

Then $\| \iota ST(x_i) - x_i\| \leq \| \iota ST(x_i) - T(x_i) \| + \| T(x_i) - x_i \| \leq \epsilon \|T(x_i)\| + \epsilon$ $\leq \epsilon^2 + \epsilon\|x_i\| + \epsilon$. So $ST : H_0 \rightarrow H$ approximates the identity in the $H_0$ norm.

• If we want a net, we just let $(x_i)_{i=1}^n$ run through all finite subsets of $H_0$, and observe that we didn't use the bound on $T$, so as Bill Johnson suggests, we could just take $T$ to be a projection onto the span of the $(x_i)$, no condition on $H_0$ needed.

• If $H_0$ is separable, let $(x_k)$ be a countable dense subset, and let $S_nT_n$ be chosen as above for $(x_i)_{i=1}^n$ and $\epsilon=1/n$. If $x\in H_0$ with $\|x - x_i\| \leq \epsilon$ for some $i\leq n$, then \begin{align*} & \| \iota S_nT_n(x) - x \| \leq \| \iota S_nT_n(x) - x \| \\ & = \| \iota S_nT_n(x-x_i) - T_n(x-x_i) + \iota S_n T_n(x_i) - T_n(x_i) + T_n(x) - x \| \\ &\leq \epsilon\|T_n(x-x_i)\| + \epsilon\|T_n(x_i)\| + \|T_n(x-x_i) - (x-x_i) + T_n(x_i) - x_i \| \\ &\leq \epsilon^2\lambda + \epsilon(\epsilon+\|x_i\|) + \epsilon\lambda + \epsilon \\ &\leq \epsilon^2\lambda + \epsilon(2\epsilon+\|x\|) + \epsilon\lambda + \epsilon. \end{align*} Without the BAP you seemingly cannot control $\|T(x-x_i)\|$ for example.

Remark 1: Having the "compact approximation property" doesn't seem to help. By definition, this means we can only choose $T$ to be compact not finite-rank. Then the image of the unit ball of $H_0$ under $T$ is a compact set, but I don't know how to form the equivalent of $S$. That is, how do you (linearly) distort a compact set from $H$ into $H_0$?