= Solution
Suppose first that the bounded <Fredholm operator> $T:H\to H$ has <Fredholm index> zero. Put $N=\ker T$ and $C=(\operatorname{ran}T)^\perp=\ker T^*$. These finite-dimensional <linear subspaces> have the same <dimension>. The restricted map $T:N^\perp\to\operatorname{ran}T$ is a bounded bijection, with bounded inverse by the <bounded inverse theorem>. Choose an isomorphism $J:N\to C$ and extend it by zero on $N^\perp$, writing the resulting <finite-rank operator> as $K$.
Relative to the <orthogonal decompositions> $H=N\oplus N^\perp$ and $H=C\oplus\operatorname{ran}T$, the operator $T+K$ is the direct sum of $J$ and the invertible restricted map. Thus $T+K$ is invertible, and $T=(T+K)-K$ is the sum of an invertible operator and a <compact operator>.
Conversely write $T=U+K$ with $U$ invertible and $K$ compact. It suffices to treat $I+C$ with $C=U^{-1}K$. A <compact operator> on a <Hilbert space> admits arbitrary <operator norm> approximation by <finite-rank operators>: take a finite epsilon-net of its unit-ball image and project onto the span of the net. Choose $F$ of finite rank with $\|C-F\|<1$, put $R=C-F$, and factor
$$
I+C=(I+R)\bigl(I+(I+R)^{-1}F\bigr).
$$
The first factor is invertible by the <Neumann series>. Set $G=(I+R)^{-1}F$ and $E=\operatorname{ran}G$, a finite-dimensional <linear subspace>. On $E\oplus E^\perp$, the second factor has block form
$$
I+G=\begin{pmatrix}A&B\\0&I\end{pmatrix},\qquad A=I_E+G|_E.
$$
Left multiplication by the invertible block matrix $\begin{pmatrix}I&-B\\0&I\end{pmatrix}$ reduces it to $A\oplus I$. Its range is closed, and its <kernel> and <cokernel> have equal <dimension> by finite-dimensional <rank-nullity theorem> for $A:E\to E$. Invertible factors preserve all these properties. Hence $T$ is a <Fredholm operator> of <Fredholm index> zero, proving both directions:
$$
\boxed{T\text{ Fredholm with index }0\iff T=U+K,\quad U\text{ invertible},\ K\text{ compact}.}
$$
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