= Solution
Fix a <Fredholm operator> $T$. Split its domain as $X\oplus N=(\ker T)^\perp\oplus\ker T$ and its target as $Y\oplus M=\operatorname{ran}T\oplus\ker T^*$. For a sufficiently small norm perturbation $R$, write
$$
T+R=\begin{pmatrix}A&B\\C&D\end{pmatrix},
$$
where $A:X\to Y$ remains invertible by the <Neumann series>. Invertible triangular row and column operations reduce this matrix to
$$
\begin{pmatrix}A&0\\0&D-CA^{-1}B\end{pmatrix}.
$$
The second block maps the finite-dimensional space $N$ into $M$. Hence the perturbed operator has closed range, finite kernel and cokernel, and
$$
\operatorname{ind}(T+R)=\dim N-\dim M=\operatorname{ind}T,
$$
since the rank of that finite block cancels from the difference. Thus \b[the <Fredholm> set is open and its index is locally constant, hence norm continuous].
If $S,T$ are <Fredholm>, their product is <Fredholm> by multiplying their invertible images in the <Calkin algebra>. For index additivity use the following finite-dimensional <exact sequence>:
$$
0\to\ker T\to\ker ST\xrightarrow{T}\ker S
\xrightarrow{q}\operatorname{coker}T\xrightarrow{S}\operatorname{coker}ST
\to\operatorname{coker}S\to0.
$$
Here $q$ is reduction modulo $\operatorname{ran}T$, the induced $S$-map sends $[x]$ to $[Sx]$, and the last map is the quotient by the larger range $\operatorname{ran}S$. These maps are well-defined. For example, $\ker q=\ker S\cap\operatorname{ran}T$ is the image of the preceding map. If $Sx\in\operatorname{ran}ST$, subtracting a suitable $Ty$ makes $x-Ty\in\ker S$, proving exactness at $\operatorname{coker}T$. The other positions follow directly from the definitions. Alternating dimensions in this exact sequence gives
$$
\boxed{\operatorname{ind}(ST)=\operatorname{ind}S+\operatorname{ind}T.}
$$
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