Universal Cover of $SL_{2}(\mathbb{R})$

Here's a slightly different way of seeing that $\DeclareMathOperator{\SL}{SL(2,\mathbb{R})}\DeclareMathOperator{\USL}{\widetilde{SL}(2,\mathbb{R})}\DeclareMathOperator{\SLC}{SL(2,\mathbb{C})}\DeclareMathOperator{\sl}{\mathfrak{sl}(2,\mathbb{R})}\DeclareMathOperator{\slc}{\mathfrak{sl}(2,\mathbb{C})}\USL$ has no faithful finite-dimensional representations:

Start with a representation $\Phi$ of $\USL$ on a finite-dimensional real vector space. The associated representation $\phi=d\Phi(1)$ of $\sl$ can be complexified to yield a representation of $\slc$. Since $\SLC$ is simply connected, this corresponds to a representation $\Psi_{\mathbb{C}}$ of $\SLC$. Since $\SL$ is a subgroup of $\SLC$, we get a representation $\Psi$ of $\SL$. The covering projection $\pi:\USL \to \SL$ yields another representation $\Psi\circ\pi$ of $\USL$. It is straightforward to check that $\phi = \psi$ ($= d(\Psi\circ\pi)(1)$). Since $\USL$ is simply connected, this means that $\Phi$ and $\Psi\circ\pi$ are actually the same representation. This shows that $\Phi$ factors over $\pi$, and hence has non-trivial kernel.

I don't have a deep answer -- just a proof.

First, I assume you want to know why $\widetilde{SL}_2(\mathbb{R})$ has no finite-dimensional faithful representations, i.e. why it cannot be expressed as a matrix group.

The proof is that the Lie algebra of $\widetilde{SL}_2(\mathbb{R})$ is $\mathfrak{sl}_2(\mathbb{R})$, so any representation of $\widetilde{SL}_2(\mathbb{R})$ must induce a representation of $\mathfrak{sl}_2(\mathbb{R})$ at the tangent space to the identity. Moreover, since $\widetilde{SL}_2(\mathbb{R})$ is simply connected, the representation of $\widetilde{SL}_2(\mathbb{R})$ is entirely determined by the representation of $\mathfrak{sl}_2(\mathbb{R})$.

Now, the representations of $\mathfrak{sl}_2(\mathbb{R})$ have been completely classified, and it is easy to check that none of them induce a faithful representation of $\widetilde{SL}_2(\mathbb{R})$. So that's that.

On a more philosophical level, there's no reason to think that every Lie group would be a matrix group, so it's not surprising that $\widetilde{SL}_2(\mathbb{R})$ turns out not to be. This happens a lot for covers of matrix groups. For example, the universal covers of the symplectic groups are the metaplectic groups, which also have no faithful finite-dimensional representations.

This comes a bit late but I think it answers shortly the question:

Theorem: Any connected linear semisimple Lie group has finite center.

(This is mainly because of Schur's Lemma)

But $Z(\widetilde{SL_2(\mathbb{R})})$ is isomorphic to $\mathbb{Z}$ the fundamental group of $SL_2(\mathbb{R})$.