An operator $T$ from a Banach space $X$ to a Banach space $Y$ is compact if and only if for every bounded sequence $x_n$ in $X$, there is a subsequence $x_{n_k}$ such that $Tx_{n_k}$ converges in $Y$ (in the norm).
Consider the adjoint operator $A^*:X^*\to (c_0)^*\simeq\ell_1$. Let $(x_n^*)$ be a bounded sequence in $X^*$. Since $X$ is reflexive, so is $X^*$. Every bounded sequence in a reflexive Banach space has a weakly convergent subsequence, so there exists a subsequence $(x_{n_k}^*)$ converging weakly.
Since $A^*$ is continuous, it is also weakly continuous, that is, $A^*x_{n_k}^*$ converges weakly to some element $y\in\ell_1$. But in $\ell_1$ weak convergence is equivalent to norm-convergence. Therefore $A^*x_{n_k}^*$ converges in norm. We have thus verified that the condition above holds, so $A^*$ is compact, and so is $A$.
Added in response to some requests for clarification.
The argument above relies on some classical theorems in Banach space theory, which apparently are not sufficiently well known.
The first theorem is that in a reflexive Banach space every bounded sequence has a weakly convergent subsequence. For a proof, see here.
The second theorem is that in $\ell_1$ the weak and norm-topologies are equivalent. In particular, weak convergence of sequences is equivalent to norm convergence of sequences. Most textbooks on Banach spaces contain this theorem. For several references, see here.
Since $\dim \ker T<\infty$ and $\mathrm{ran}\,T$ is closed, it results that $T$ is semi-Fredholm. Hence $T^*$ is semi-Fredholm with $\mathrm{codim}\,\mathrm{ran}\,T^*<\infty$ (and $\mathrm{ran}\,T^*$ is closed). Furthermore we have
$$(1) \quad \mathrm{ind}\,T=-\mathrm{ind}\,T^*.$$
Now let $K:= T -T^*$. Since $K$ is compact, $T-K$ is semi- Fredholm with
$$(2) \quad \mathrm{ind}\,T=\mathrm{ind}\,(T-K).$$
But we have $T^*=T-K$, hence
$$(3) \quad \mathrm{ind}\,T=\mathrm{ind}\,T^*.$$
From $(1)$ and $(3)$ we now see, that $ \mathrm{ind}\,T=\mathrm{ind}\,T^*$ is finite $=0$ and therefore $\mathrm{codim}\,\mathrm{ran}\,T<\infty.$
Best Answer
By compactness of $K$, the closed unit ball in $\ker(I+K)$ is compact, so $\ker(I+K)$ has finite dimension. Therefore, $X = \ker(I+K) \oplus S$ for some closed subspace $S$ of $X$. The operator $I + K$ restricted to $S$ has an inverse $T : \operatorname{Im}(I+K) \to S$. You can show (by way of contradiction, using compactness of $K$) that $T$ is continuous at $0$ (and hence continuous). Then $\operatorname{Im}(I + K)$ is closed.
To demonstrate that $\operatorname{Im}(I + K)$ has finite codimension, suppose to the contrary that the codimension of $\operatorname{Im}(I + K)$ is infinite. There is a sequence $\operatorname{Im}(I + K) = S_0 \subsetneq S_1 \subsetneq S_2 \subsetneq \cdots$ of closed subspaces of $X$ with $\dim(S_{n+1}/S_n) = 1$ for every $n$. Riesz's lemma gives normalized vectors $x_n\in S_n$ such that $\operatorname{dist}(x_n,S_{n-1}) \ge 0.5$. If $n > m$, then $(I + K)x_n, (I + K)x_m$, and $x_m$ belong to $S_{n-1}$, forcing $$\|Kx_n - Kx_m\| = \|(I + K)x_n - (I + K)x_m + x_m - x_n\| \ge \operatorname{dist}(x_n,S_{n-1}) \ge 0.5$$ Consequently, $(Kx_n)$ has no convergent subsequence, contradicting compactness of $K$.