Let $R$ be a commutative Banach algebra.
Let $E_n$ denote the group generated by $n\times n$ elementary matrices. $SK_1(R)$ denotes the kernel of the induced determinant map $K_1(R) \to R^{\times}$.
In the K-book, Weibel states,
$E_n(R)$ is the path component of the identity matrix $I_n$ for any $n\geq 2$. Hence we may identify the $SK_1(R)$ with the set $\pi_0{SL}(R)$.
I do not see how the second statement follows from the first. I am just looking for a brief explanation that helps me make sense of this lemma.
You're interested in $K_1(R) = GL_\infty(R)/E_\infty(R)$, where these are defined as the colimits of the corresponding terms $GL_n(R)$ and $E_n(R)$.
What you know is that $E_\infty(R)$ is a normal subgroup of $GL_\infty(R)$ that is contained inside $SL_\infty(R) = \text{ker}(\det)$ (another normal subgroup). Now the third isomorphism theorem gives $$\left(GL_\infty(R)/E_\infty(R)\right)/\left(SL_\infty(R)/E_\infty(R)\right) = GL_\infty(R)/SL_\infty(R);$$ rephrasing, there is a short exact sequence $$SL_\infty(R)/E_\infty(R) \to GL_\infty(R)/E_\infty(R) \to GL_\infty(R)/SL_\infty(R).$$
The second term is $K_1(R)$ and the last term is $R^\times$, both by definition. But now you know from Weibel's proposition 1.5 (the one that you quote) that the first term is $\pi_0 SL_\infty(R)$, because for any topological group $G$, the quotient $G/G_0$ by the identity component is naturally identified with $\pi_0 G$. You also know the second term is $R^\times$. So you have a short exact sequence $$\pi_0 SL_\infty(R) \to K_1(R) \to R^\times,$$ where the last map is the determinant. Of course, the kernel of the determinant map $K_1(R) \to R^\times$ is precisely how $SK_1(R)$ is defined, so we have the desired result.