If $\|p-q\|<{1\over2}$ then $p$ is homotopy equivalent to $q$ 
Let $A$ be a $C^*$ algebra, $p,q \in A$ projections, such that
  $\|p-q\|< {1 \over 2}$. Show that $p$ homotopy equivalent to $q$.

Proof. Let $a_t=(1-t)p+tq$, then $a_t$ is positive (self-adjoint with positive spectrum) and $||a_t||\leq1.$ Moreover, $\|a_t-p\|<{t\over2}$ and $a_1=q$ so $\|a_t-p\|<{1\over2}$ for all $t\in [0,1]$. It is not hard to show that ${1\over2}\notin\sigma(a_t)$. Define $f:[0,1]\to [0,1]$ by $f(x)=\chi_{[{1\over2},1]}$, then $f\in C(\sigma(a_t))$ is a homotopy of projections, as required.
I can't show that $t\to f(a_t)$ is continuous. Here is my attempt for $t_0=0$, please check I did it correctly:
Let ${1\over2}>\varepsilon>0$ be given. If $t<\varepsilon$ then $\|a_t-p\|<{t\over2}<{\varepsilon\over2}$. 
Claim: If $\lambda\in [\varepsilon, 1-\varepsilon]$ then  $ \lambda \notin \sigma(a_t)$. Indeed, $$\|(a_t-\lambda)-(p-\lambda)\|<{\varepsilon\over2}$$ and $\sigma(p)\subseteq\{0,1\}$ implies $\sigma(p-\lambda)\subseteq\{-\lambda,1-\lambda\}$;  in particular, it  doesn't contain $0$, so $p-\lambda$ is invertible.
Now, $\sigma((p-\lambda)^{-1}) \subseteq \{-{1\over\lambda}, {1\over{1-\lambda}}\}$ and it is self adjoint, so $\|(p-\lambda)^{-1}\|\in\{{1\over\lambda}, {1\over{1-\lambda}}\}$. Then $$\|(p-\lambda)^{-1}\|^{-1}\in\{\lambda, 1-\lambda\},$$ so $$\|(p-\lambda)^{-1}\|^{-1}\ge \varepsilon > {\varepsilon \over 2}.$$
Thus $\|(a_t-\lambda)-(p-\lambda)\|<\|(p-\lambda)^{-1}\|^{-1}$ implies $a_t-\lambda$ is invertible and $\lambda \notin \sigma(a_t)$. 
So, $\sigma(a_t) \subseteq [0,\varepsilon)\cup (1-\varepsilon,1]$ and we get 
$$\|f(a_t)-a_t\|=|(f(x)-x)|_{\sigma(a_t)}||_\infty<\varepsilon$$ by Gelfand transform, and $$\|f(a_t)-f(a_0)\|\leq\|f(a_t)-a_t\|+\|a_t-p\|<1.5\varepsilon,$$ fair enough.
I don't know if my proof for continuity in zero is O.K, and don't know how to prove the general case. 
Thank you in advance.
 A: Your proof is fine. What you are missing to work at points other than zero is the following lemma:
Lemma. Let $f:X\to\mathbb C$ with $X$ a compact subset of $\mathbb R$, $\varepsilon>0$ and $R>0$. Then there exists $\delta=\delta(\varepsilon,R)>0$ such that if $a,b\in A^+$, with $\|a\|+\|b\|<R$, with $\sigma(a)\cup\sigma(b)\subset X$, and such that $\|a-b\|<\delta$, then $\|f(a)-f(b)\|<\varepsilon$. 
Proof. Let $p(t)=\sum_{j=0}^n r_j\,t^j$ be a polynomial such that $\|f-p\|_\infty<\varepsilon/3$ on $X$. Note that, using the equality
$$
t^j-s^j=(t-s)\,\sum_{k=0}^{j-1}t^ks^{j-1-k},
$$ we have 
$$
\|p(a)-p(b)\|\leq\sum_{j=1}^n|r_j|\,\|a^j-b^j\|\leq \left(\sum_{j=1}^n\sum_{k=0}^{j-1}|r_j|\,\|a\|^k\|b\|^{j-1-k}\right)\,\|a-b\|
\leq L\,\|a-b\|,
$$ where
 $$ L=\sum_{j=1}^n\sum_{k=0}^{j-1}|r_j|\,R^k\,R^{j-1-k}=\sum_{j=1}^n\sum_{k=0}^{j-1}|r_j|\,R^{j-1}=\sum_{j=1}^n\,j\,|r_j|\,R^{j-1}.$$ Now put $\delta=\varepsilon/3L$. Then, if $\|a-b\|<\delta$,
\begin{align}
\|f(a)-f(b)\|&\leq\|f(a)-p(a)\|+\|p(a)-p(b)\|+\|p(b)-f(b)\|\\ \ \\
&\leq \|f-p\|_\infty+L\,\|a-b\|+\|p-f\|_\infty\\ \ \\
&\leq\frac\varepsilon3+\frac\varepsilon3+\frac\varepsilon3=\varepsilon.& \Box
\end{align}
Now the proof is straightforward: given $\varepsilon>0$, choose $\delta$ from the Lemma with $R=1$ and $X=[-1/2,3/2]$. If $|t-s|<\delta$, then 
$$
\|a_t-a_s\|=\|(t-s)(p-q)\|\leq|t-s|<\delta,
$$
and so 
$$
\|f(a_t)-f(a_s)\|<\varepsilon.
$$
So $t\longmapsto f(a_t)$ is continuous.
