Showing higher homotopy groups of $S^1$ are trivial I'm trying to prove $\pi_{i} (S^1) \cong 0$ if $i>1$.
Is this correct. 
You have a short exact sequence,
$\mathbb{Z} \rightarrow \mathbb{R} \rightarrow S^1$ (from the fiber bundle of the covering space), then you deduce from some magic I don't understand you can do this
$\pi_{i}(\mathbb{Z}) \rightarrow \pi_{i}(\mathbb{R}) \rightarrow \pi_{i}(\mathbb{R}) \rightarrow\pi_{i-1}(\mathbb{Z}) \rightarrow \ldots$
Is this correct? 
I was wondering about $\pi_{i}(\mathbb{R})$ being trivial when $i>1$. But, yeah made it unclear as I was asking two questions. 
 A: Just as a remark - and in the end it amounts to Jason DeVito's answer - but you don't need the full power of the long exact sequence of a fibration to prove this - just general covering space theory. 
In fact it is a general fact (which does however easily follow from the LES) is that for $n \ge 2$ a covering $\tilde{X} \to X$ induces an isomoprhism $\pi_n(\tilde{X}) \to \pi_n(X)$. The point is you don't need to think about fibrations if you aren't familiar with them (see Prop 4.11 in Hatcher for details). 
A: This is not the standard way to prove the higher homotopy groups of $S^1$ are trivial, but I'll bite.
First, a series of lemmas.
Lemma 0: Each connected component of $\mathbb{Z}$ (i.e, each point) is contractible - the identity map, restricted to each component, is homotopic to a constant map.
Proof:  The connected components are points, so the identity map is constant.
Lemma 1:  $\mathbb{R}^n$ is contractible, that is, the identity map is homotopic to a constant map.
Proof:  Consider the homotopy $F(x,t) = tx$.  When $t=0$, this is a constant map and when $t=1$, it's the identity map.  (I'll leave it to you to prove $F$ is continuous).
Lemma 2:  If $X$ is contractible, then all homotopy groups are trivial.
Proof:  Given a map $g:S^n\rightarrow X$, we need to find a homotopy between $g$ and a constant map.  Well, let $F$ be a homotopy witnessing the contractibility of $X$.  Then the map $G:S^n\times I$ defined by $G(x,t) = F(g(x),t)$ has the desired properties.
Putting this altogether shows that $\pi_i(\mathbb{Z}) = \pi_i(\mathbb{R}) = \{e\}$ for $i > 0$ (where we pick a component of $\mathbb{Z}$ once and for all).  Can you take it from there?
A: You can lift the sphere map $f: S^k \to S^1$ through the universal covering space  $\pi: \mathbb R\to  S^1$ since 
$1=f_*\pi_1(S^k)\leq\pi_1(S^1)=\mathbb Z$ and contract inside $\mathbb R$ for $k\geq 2$. I am not sure you can do HES through disconnected spaces. 
