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.
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.