Does every Banach space admit a continuous injection to a non-closed subspace of another Banach space?

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Let $(V, \lVert\,\rVert)$ be a Banach space. I want to produce a non-complete norm $\lVert\,\rVert'$ on it such that $\lVert v\rVert' \leq \lVert v\rVert$ for all $v$ in $V$. Given a continuous injection $\varphi\colon V \to W$ with a non-closed image, taking $\lVert v\rVert' = C \lVert \varphi(v)\rVert$ for some $C$ provided by continuity (preimage of a unit open ball from $W$ is open in $V$, hence contains a smaller ball, so we have an inequality of norms) results in an incomplete norm. However, I wasn't able to find a simple (and non-circular) way to map a space into a non-closed subspace.

The question is inspired by a proof of the fact that $V$ is homeomorphic to $V \backslash K$ for all compact subsets $K$ from a book "Selected Topics in Infinite-Dimensional Topology" (C. Bessaga, A. Pelczynski). The existence of a non-complete norm is an essential part of the proof. In the book it is obtained first for separable spaces (by embedding into $l^2$ as a non-closed subspace) and then in general by combining the new norm on a separable subspace and old norm on the entire space.

That proof is short and quite simple, but not particularly elegant. It would be nice to have a uniform treatment for separable and non-separable spaces. (And it would be really neat if every Banach space could inject as a proper non-closed subspace of itself, but that doesn't sound likely).