Just to add an answer which explicitly uses the [Approximation property](https://en.wikipedia.org/wiki/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.