Let $X$ be a separable topological vector space with size (cardinal number) no larger than $\mathfrak{c}$. Does there exist any sequence of finite rank linear maps $\phi_n:X\to X$ pointwise converging to the identity mapping $id:X\to X$?

$\begingroup$ Do you know whether this is true for, say, Banach spaces? $\endgroup$ – Jochen Glueck May 31 '18 at 14:21

1$\begingroup$ I think that a separable Banach space without the approximation property would be a counterexample. I'll try to write down a proof in a couple of hours, but I wonder if there is something easier. $\endgroup$ – Nate Eldredge May 31 '18 at 14:27
Just to chat, an easier counterexample is $X:=L^p(\mathbb{R})$ for $0\le p<1$, a complete metric separable TVS. The identity map can't be approximated by finite rank continuous linear operators, for the simple reason that there aren't any. The only convex open set of $X$ is $X$ itself. As a consequence, there aren't any nonzero continuous linear forms on $X$, nor any nonzero finite rank continuous linear operators.
I presume you want the operators $\phi_n$ to be continuous.
Let $X$ be a separable Banach space (which necessarily has cardinality $\mathfrak{c}$). If such a sequence exists, then $X$ has the approximation property. To see this, first note that we have $\phi_n \to id$ uniformly on compact subsets of $X$ (this is a nice exercise in the uniform boundedness principle  a good "prelim" problem). Now if $A : X \to X$ is a compact operator, it maps bounded sets inside compact sets, so $\phi_n A \to A$ uniformly on bounded sets, i.e. in operator norm. Thus $A$ is a norm limit of finiterank operators.
However, not every separable Banach space has the approximation property. The first counterexample was constructed by Enflo in 1973. See the abovelinked Wikipedia article for references.
It's quite possible that there is an easier counterexample.