Let $\tilde a,\tilde b:I\rightarrow X$ be lifts of $a,b:I\rightarrow Y$ satisfying $p\circ \tilde a=a$, $p\circ \tilde b=b$, and let $F_t:a\simeq_{\partial I}b:I\rightarrow Y$ be a relative homotopy satisfying $F(s,0)=a(s)$, $F(s,1)=b(s)$, $F(0,t)=a(0)=b(0)$ and $F(1,t)=a(1)=b(1)$. Now apply the HELP lemma in the diagram
$\require{AMScd}$
\begin{CD}
0\times I\cup I\times \partial I@>G'>> X\\
@Vi V V @VVpV\\
I\times I @>F>> Y
\end{CD}$\require{AMScd}$
where $G'(s,0)=\tilde a(s)$, $G'(s,1)=\tilde b(s)$, $G'(0,t)=\tilde a(0)=\tilde b(0)$. Note that $p\circ G'=F\circ i$. From the data the HELP lemma gives a map $G:I\times I\rightarrow X$ satisfying
$$G(s,0)=\tilde a (s),\quad G(s,1)=\tilde b(s),\quad G(0,t)=\tilde a(0)=\tilde b(0),\quad p\circ G(s,t)=F(s,t).$$
In particular $p(G(s,1))=p(\tilde a(1))=p(\tilde b(1))=F(s,1)=a(0)=b(0)$, so the assignment $s\mapsto G(s,1)$ is a path in $X$ contained in the fibre over $a(0)=b(0)$, going from $\tilde a(0)$ and ending at $\tilde b(0)$. Since the covering projection has discrete fibres such a path must be constant.
Hence $\tilde a(1)=\tilde b(1)$, and $G:\tilde a\simeq_{\partial I}\tilde b$ is a relative homotopy.
Solution 1: Observe that $S^1\times [0,1]$ is compact. So the lift of the null-homotopic map, lets call it $\bar{f}$, has its image in $[-n,n]\times [0,1]$ for some $n$. (Also I am assuming that the covering map $\mathbb R\to S^1$ is $x\mapsto e^{2\pi i x}$.) Thus we can think of $\bar{f}$ as a map from $[0,1]\times[0,1]\to [-n,n]\times [0,1]$. Also observe that $\bar{f}$ restrict onto $\{0\}\times [0,1]$ and $\{1\}\times [0,1]$ are identical. So infact you can extend $\bar{f}: [-n,n]\times [0,1]\to [-n,n]\times [0,1]$. And both of them are homeomorphic to $D^2$. So by Brower fixed point theorem, it has a fixed point. And now look at the image of this fixed point under the covering projection, and that will be a fixed point for $f$.
Solution 2: Use Lefschetz fixed point theorem (https://en.wikipedia.org/wiki/Lefschetz_fixed-point_theorem). Since $f$ is null homotopic, $f_*$ induces zero map on $H_1(S^1\times[0,1])$. Also since $S^1\times [0,1]$ is homotopic to $S^1$ thus $H_2=0$. And $f_*$ induces identity map on $H_0$. So $\Lambda_f=1\neq 0$. Thus have a fixed point.
Best Answer
You don't need the Brouwer fixed point theorem, you just need a very concrete construction.
The key fact is that for every two points $P,Q \in T^2$ there exists a homeomorphism $g : T^2 \to T^2$ such that $g(P)=Q$ and such that $g$ is homotopic to the identity. Once you know that, choosing any $Q \in T^2$ and letting $P = f(Q)$, it follows that $g \circ f$ is homotopic to $f$ and $g \circ f(Q) = Q$.
The function $g$, and the homotopy, can be defined first up in the universal cover. Choose lifts $\tilde P, \tilde Q \in \mathbb R^2$ of $P,Q$. Now just do a straight line homotopy, sliding each point along the displacement vector $\tilde Q - \tilde P$: $$\tilde H(a,t) = a + (1-t)(\tilde Q-\tilde P) $$ and so $\tilde g(\tilde P)=\tilde H(\tilde P,0)=\tilde Q$, and $\tilde H(\tilde P,1)=\tilde P$.
Now you have to show that this straight line homotopy upstairs induces a homotopy downstairs. More precisely, letting $[a]=a+\mathbb Z^2 \in T^2$ denote the orbit of $a$ under the action of the deck group, you have to check that the formula $$H([a],t) = [a+(1-t)(\tilde Q- \tilde P)] $$ gives a well-defined homotopy on $T^2$, and that the function $g([a]) = H([a],0)$ satisfies $g(P)=Q$, and $g([a],1)=[a]$ for all $[a] \in T^2$.