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Write $A=T-\lambda I$. We first prove that $\mathrm{im}A$ is dense. Otherwise some nonzero $y$ lies in its orthogonal complement, soBy part (b), choose finite-dimensional spaces $E_j$ containing $y$ such that $S_j=P_{E_j}T\to T$ in operator norm. Sincethe operator $S_j^{\ast }-\overline{\lambda }I$ 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 $S_j{x}_j=\lambda x_j$. ThenCompactness gives a subsequence for which $T{x}_j$ converges. Since $\lambda \ne 0$, the displayed relation makes $x_j$ converge to a unit vector $x$, and continuity gives $Tx=\lambda x$, contradicting the hypothesis. Thus $\mathrm{im}A$ is dense.

Next suppose $A$ were not bounded below, or equivalently that its lower norm were zero. There would be unit vectors $x_j$ with $A{x}_j\to 0$, henceAgain a convergent subsequence of $T{x}_j$ forces the corresponding $x_j$ to converge to a unit vector $x$ satisfying $Ax=0$, another contradiction. Therefore some $c>0$ satisfies

This lower bound makes $\mathrm{im}A$ closed: if $A{x}_j$ 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.