There is no need of the use of degree. Just apply the covariant functor $H_{n-1}(-)$ and you will get your desired contradiction. More precisely on homology we would have
$$\begin{array}{ccccc} &H_{n-1}(S^{n-1}) &\stackrel{ i_\ast}{\longrightarrow} &H_{n-1}(D^n)& \stackrel{r_\ast}{\longrightarrow} & H_{n-1}(S^{n-1}) \\
\end{array}$$
which are respectively (left to right) $\Bbb{Z}$, 0, $\Bbb{Z}$ contradicting the fact that the identity map factors through zero. I should add that this is one of the standard proofs of the Brouwer fixed point theorem. For example Rotman does it this way in his book on Algebraic Topology, while Hatcher proves it in the case $n=2$ by applying the functor $\pi_1(-)$ instead of homology.
If you want a proof that uses degree, it is not so hard. Suppose you have a map $f:D^n \to D^n$ that has no fixed points. Then we can treat $f$ as a map from the northern hemisphere $D^n_+$ of $S^n$ to itself. Now we can extend $f$ to a map on $S^n$ as follows. We define
$$g(x) = \begin{cases} f(x), & \text{if}\hspace{2mm} x \in D^n_+,\\
f\circ r(x), & \text{if}\hspace{2mm} x \in D^n_{-} \end{cases}$$
where $r(x)$ is reflection about the plane through the equator and $D^n_{-}$ is the southern hemisphere. It is clear that $g(x)$ is a continuous function; furthermore $g(x)$ has no fixed points. Hence we can homotope $i\circ g$ to the antipodal map on $S^n$ that has degree $(-1)^{n+1}$, where $i : D^n_{+} \hookrightarrow S^n$ is inclusion.
However $i \circ g$ is not surjective because for example no point in the southern hemisphere is in the image. It follows that $\deg i \circ g = 0$ contradicting the fact that we found it to be $(-1)^{n+1}$.
Fundamental groups are not much use for dealing with disconnected spaces since the group cannot see anything outside the component of the base-point. For this case where you want to show there is no surjective $[-1,1] \to \{-1,1\}$ we can use the much cruder topological notion of connectedness. The above map does not exist because the domain is connected and the range isn't.
If you are set on using an algebraic invariant then you can use the $0$th homology group which is just the free group over the number of path components. Or you can use the fundamental groupoid which also contains somehow the number of path components and, unlike the fundamental group, can see all (path) components.
Best Answer
Yes. Let $f\colon [a,b]\to \mathbb R$ be continuous, and assume, without loss of generality, that $f(a)<f(b)$. Let $t\in \mathbb R$ with $f(a)<t<f(b)$. Assume that $t$ is never attained by $f$.
Construct $g\colon [a,b]\to \mathbb R$ by:
We don't have to define $g$ on points in which $f(x)=t$, since $f$ never equals $t$ in $[a,b]$.
Then $g$ is a continuous function whose image is contained in $[a,b]$. By the assumption on $t$, $g(a)=b$, and $g(b)=a$, and $g(x)\ne x$ for all $x\in (a,b)$ (by construction of $g$). In other words, $g$ has no fixed-point, contradicting Brouwer's fixed-point theorem.