Multiple of identity plus compact

OK, let me try too. It is going to be a somewhat long story. WLOG, $\|T\|\le 1$.

Step 1: It is enough to show that for every finite-dimensional subspace $E$ and every $\delta>0$, there exists a unit vector $v\in E^\perp$ and $z\in\mathbb C$ such that $\|Tv-zv\|\le\delta$.

Proof: Suppose that the claim holds. Choose a square summable sequence $\delta_j>0$. Construct by induction a sequence of pairwise orthogonal unit vectors $v_j$ and numbers $z_j$ such that $\|Tv_j-z_jv_j\|\le\delta_j$. Passing to a subsequence, we can ensure that $z_j\to z$ and, moreover, $\sum_j|z_j-z|^2<+\infty$. Then $T-zI$ is a Hilbert-Schmidt operator on $\operatorname{span}(v_j)$.

Step 2: Suppose that $1>\delta>0$ and $a_j$ is a sufficiently long (i.e., of length $\ge N(\delta)$) sequence of positive numbers such that $\delta^2 a_j\le a_{j+1}\le a_j$. Then there exist $m<n$ such that if we multiply $a_j$ by an (automatically non-decreasing) geometric progression so that the resulting products $a_j'$ will satisfy $a_m'=a'_n=1$, then we shall have $\sum_{k=m}^{n-1} a_k'\ge \delta^{-2}$.

Proof: Consider the points $P_k=(k,\log a_k)$, $k=0,\dots,N$ and take the convex hull of rays going from those points up. It will be bounded by two vertical rays and several slanted intervals. Now, if there is a slanted interval between $P_m$ and $P_n$ with $n-m>\delta^{-2}$, we are done (because for the corresponding modified sequence $a_k'$, we'll have $a_k'\ge 1$ for $m\le k\le n$). Otherwise, since the slopes of slanted intervals are squeezed between $2\log\delta$ and $0$ and increasing, we can find $2\delta^{-2}$ subsequent intervals that form an almost straight piece, provided that $N$ is large enough. Then if $P_m$ is the left endpoint of the left of those intervals and $P_n$ is the right endpoint of the right of those intervals, we have $n-m\ge 2\delta^{-2}$ and $a_k'\ge\frac 12$ for $k$ between $m$ and $n$.

Step 3: Let $E$ be a finite-dimensional subspace. Let $\delta>0$ and let $N$ be large enough to guarantee the conclusion of Step 2. The conditions $x,Tx,T^2x,\dots, T^Nx\in E^\perp$ define a closed subspace of finite codimension in our infinite-dimensional Hilbert space $H$. Let $x$ be a unit vector in that subspace. Consider the sequence $x_k=T^kx$, $k=0,\dots, N$ and the numbers $a_k=\|x_k\|^2$. If we have $a_{j+1}<\delta^2a_j$, then the normalized vector $x_j$ is what we are looking for with $z=0$. Otherwise choose $m<n$ as in Step 2. Let $q^2\ge 1$ be the ratio of the corresponding geometric progression and $r^2$ be its $m$-th term. Then for the vectors $y_k=rq^{k-m}x_k$, we have $\|y_m\|=\|y_n\|=1$ and $y_{k+1}=qTy_k$. Let $|\zeta|=1$. Consider the vector $$ y=y_\zeta=\sum_{k=m}^{n-1}\zeta^{-k}y_k. $$ Then $\|Ty-\zeta q^{-1}y\|=\frac 1q\|\zeta^{-n}y_n-\zeta^{-m}y_m\|\le 2$. However, the average of $\|y_\zeta\|^2$ is $\sum_{k=m}^{n-1}\|y_k\|^2\ge \delta^{-2}$, so we can use the normalized $y_\zeta$ with appropriately chosen $\zeta$ (with $2\delta$ instead of $\delta$).

It is the last step that uses the Hilbert space structure in a really essential way (though the previous steps made some limited use of it too), which makes me wonder what happens in an arbitrary Banach space.

I think this is essentially Theorem 3.1 in:

A. Brown, C. Pearcy, Compact restrictions of operators, Acta. Sci. Math (32) 1971, 271-282 (link)

In fact, something a bit stronger and in more generality is true (see Proposition 3.2 in this paper):

Proposition: Let $X$ is a Banach space and $T\in\mathcal{B}(X)$ such that there exists $\lambda\in\partial\sigma(T)\setminus\sigma_p(T)$. Then there exists $e\in X$ such that for any $\epsilon>0$ we can find $Y$ an infinite dimensional and infinite co-dimensional subspace of $X$ with the property that $(\lambda I-T)_{|Y}:Y\to Y+[e]$, and $(\lambda I-T)_{|Y}$ is compact and has norm less than $\epsilon$.