= Solution
Write $A=T-\lambda I$. We first prove that $\operatorname{im}A$ is dense. Otherwise some nonzero $y$ lies in its <orthogonal complement>, so
$$
A^*y=0,
\qquad T^*y=\overline\lambda y.
$$
By part (b), choose finite-dimensional spaces $E_j$ containing $y$ such that $S_j=P_{E_j}T\to T$ in <operator norm>. Since
$$
S_j^*y=T^*P_{E_j}y=T^*y=\overline\lambda y,
$$
the operator $S_j^*-\overline\lambda I$ is not injective. Therefore $S_j-\lambda I$ is not surjective; the contrapositive of part (c) makes $\lambda$ an eigenvalue of $S_j$. Choose unit vectors $x_j$ with $S_jx_j=\lambda x_j$. Then
$$
\|Tx_j-\lambda x_j\|
\leq\|T-S_j\|\longrightarrow0.
$$
Compactness gives a subsequence for which $Tx_j$ converges. Since $\lambda\ne0$, the displayed relation makes $x_j$ converge to a unit vector $x$, and continuity gives $Tx=\lambda x$, contradicting the hypothesis. Thus $\operatorname{im}A$ is dense.
Next suppose $A$ were not bounded below, or equivalently that its <lower norm of an operator>[lower norm] were zero. There would be unit vectors $x_j$ with $Ax_j\to0$, hence
$$
Tx_j=\lambda x_j+o(1).
$$
Again a convergent subsequence of $Tx_j$ forces the corresponding $x_j$ to converge to a unit vector $x$ satisfying $Ax=0$, another contradiction. Therefore some $c>0$ satisfies
$$
\|(T-\lambda I)x\|\geq c\|x\|.
$$
This lower bound makes $\operatorname{im}A$ closed: if $Ax_j$ converges, then $x_j$ is Cauchy and its limit maps to the same limit. Since the image is both dense and closed, it is all of $H$. We have proved the <Fredholm alternative for a compact operator>: every nonzero spectral value of a compact operator is an eigenvalue.
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