Definition: Suppose $E$ is a subspace of normed space $X$. Then $E$ is approximately complemented in $X$ if for any compact subset $K$ of $E$ and any $\epsilon>0$ there is a continuous linear operator $P\colon X\to E$ such that $\|x-P(x)\|<\epsilon$ for all $x\in K$.
Question: Is there any subspace of a Hilbert space which is not approximately complemented?
The answer is no. That is, any subspace $E$ of a Hilbert space $X$ is approximately complemented. Indeed, take any compact subset $K$ of $E$ and any real $\epsilon>0$.
Let points $x_1,\dots,x_n$ in $K$ form an $\epsilon/2$-net of $K$, and then let $P$ be the orthogonal projector from $X$ onto the linear span of $x_1,\dots,x_n$, so that $P$ is a continuous linear operator of norm $\le1$, which may be considered as a map from the Hilbert space $X$ to $E$. Take now any $x\in K$. Then $\|x-x_i\|<\epsilon/2$ for some $i=1,\dots,n$. Hence,
$$\|x-Px\|\le\|x-x_i\|+\|x_i-Px_i\|+\|P(x_i-x)\|<
\epsilon/2+0+\epsilon/2=\epsilon. $$
Iosif Pinelis has given a direct argument, but if you are studying approximate complementation in more general Banach spaces then the following argument might be of interest.
In Zhang's original paper (Proc. Amer. Math. Soc. 127 (1999), 3237-3242) he remarks that if $E$ is a closed subspace of $X$ and $E$ has the approximation property then it is approximately complemented in $X$. Although the approximation property does not always pass to subspaces, when $X$ is a Hilbert space $E$ is also a Hilbert space and hence $E$ will have the approximation property.
