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 $\|Ty-\zeta q^{-1}y\|=\frac 1q\|y_n-y_m\|\le 2$. 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.

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.

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 $q\ge 1$ be the ratio of the corresponding geometric progression and $r$ 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 $\|Ty-\zeta q^{-1}y\|=\frac 1q\|y_n-y_m\|\le 2$. 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.

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 $\|Ty-\zeta q^{-1}y\|=\frac 1q\|y_n-y_m\|\le 2$. 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.