Indeed, the proof in the linked paper does not seem valid.

Here is a "real" proof of the lemma in question, without using complex analysis:

Let $m:=$maxroot and $q_t:=(1-t)p+tq$, so that $q_0=p$ and $q_1=q$.
As in your post, without loss of generality (wlog) $m(p)\le m(q)$. If $m(p)=m(q)$, then $m(p)$ is a common root of both $p$ and $q$, and hence $m(p)$ is a root of $q_t$ for all $t$, so that $m(q_t)\ge m(p)$, as desired. So, wlog $m(p)<m(q)$, and then, by shifting and rescalling, wlog
\begin{equation*}
m(p)=0,\quad m(q)=1.
\end{equation*}
We want to show that
\begin{equation*}
m(q_t)\ge0
\end{equation*}
for all $t\in[0,1]$.
Let
\begin{equation*}
t_*:=\min\{t\in[0,1]\colon m(q_s)\ge0\ \forall s\in[t,1]\}.
\end{equation*}
This $\min$ exists in view of the continuity of $q_s$ in $s$ and the compactness of $[0,1]$.

Moreover, for each $t\in[0,t_*]$, $q_t$ is a convex combination of $p=q_0$ and $q_{t_*}$. So, wlog $t_*=1$, so that
\begin{equation*}
m(q_{t_n})<0 \tag{1}
\end{equation*}
for some sequence $(t_n)$ such that $t_n\uparrow1$.

One of the following two cases must occur:

*Case 1: there is a root $r\in(0,1)$ of $q$ of an odd multiplicity.* Then $q_1=q$ changes its sign at $r$. So, for all $t\in(0,1)$ close enough to $1$, $q_t$ will change its sign in some small enough neighborhood of $r$ and thus
will have a root close enough to $r$, which will contradict (1).

*Case 2: there is no root $r\in(0,1)$ of $q$ of an odd multiplicity.* Then $q$ does not change its sign on $(0,1)$ from $+$ to $-$ or vice versa. Therefore and because $q(0)>0$ (as was shown in your post), we see that $q\ge0$ on $[0,1]$ (and $q>0$ on $(1,\infty)$). Moreover, $1$ is a root of $q$.
Also, $q_t>0$ on $[0,\infty)$ for all $t>0$, because $p>0$ on $(0,\infty)$. So, switching from $q$ to
$q_t$, we will lose the (multiple) root $1$ of $q$, as well as all the other roots of $q$ in $(0,\infty)$ (all of them of an even multiplicity). Moreover, if $t$ is close enough to $1$, this loss cannot be regained elsewhere. Thus, we get a contradiction with the assumption that $q_t$ is stable (that is, with the assumption that all roots of $q_t$ are real).

**Details on the loss and no gains in Case 2:** We have $q_t/t=q+sp$, where $s:=(1-t)/t$ and $t:=t_n$. So, $t\uparrow1$ and hence $s\downarrow0$. For a real polynomial $P$ and an interval $I\subseteq\mathbb R$, let $N_I(P)$ denote the number of roots of $P$ in $I$ (counting the multiplicities), so that $N_{\mathbb R}(P)$ is the number of real roots of $P$; moreover, $N_{\mathbb R}(P)$ equals the degree of $P$ if $P$ is stable.

In view of (1), $N_{(0,\infty)}(q+sp)=0$, whereas $N_{(0,\infty)}(q)\ge2$ (this is what was referred to as the loss). On the other hand, the condition that the $q_t$'s are stable monic polynomials of the same degree implies that $N_{\mathbb R}(q+sp)=N_{\mathbb R}(q)$. So, to get a contradiction, it suffices to show that
\begin{equation*}
N_{(-\infty,0]}(q+sp)\le N_{(-\infty,0]}(q) \tag{2}
\end{equation*}
for all small enough $s>0$ (this is what was referred to as the no-gain).

By standard and easy bounds on the roots of a polynomials and by compactness, eventually (that is, for all small enough $s>0$) every root of $q+sp$ will be in arbitrarily small neighborhood of a root of $q$. Let $v$ be any (necessarily real) root of $q$, so that
\begin{equation*}
q(x)=(x-v)^l q_1(x)
\end{equation*}
for all real $x$, where $l$ is the natural number equal the multiplicity of the root $v$ of $q$, so that $q_1$ is a polynomial with $b:=q_1(v)\ne0$.
Let then
\begin{equation*}
p(x)=(x-v)^k p_1(x)
\end{equation*}
for all real $x$, where $k$ is the unique nonnegative integer such that $p_1$ is a polynomial with $a:=p_1(v)\ne0$.

One of the following two cases must occur:

*Case 2.1: $l\le k$.* Then, with $u:=x-v$,
\begin{equation*}
q(x)+sp(x)=u^l(q_1(x)+su^{k-l}p_1(x))\sim bu^l
\end{equation*}
as $s\to0$ and $x\to v$, so that $N_V(q+sp)=l=N_V(q)$ for some neighborhood $V$ of $v$ and all small enough $s>0$.

*Case 2.2: $l>k$.* Then, again with $u:=x-v$,
\begin{equation*}
q(x)+sp(x)=u^kp_1(x)\big(R(u)u^{l-k}+s\big),
\end{equation*}
where
\begin{equation*}
R(u):=\frac{q_1(v+u)}{p_1(v+u)}.
\end{equation*}
Note that $(R(u)u^{l-k})'=R(u)(l-k)u^{l-k-1}+R'(u)u^{l-k}\sim\dfrac ba\,(l-k)u^{l-k-1}$ as $u\to0$. So, $R(u)u^{l-k}$ is monotonic with nonzero derivative in $u$ in a right neighborhood of $0$ and in a left neighborhood of $0$. It follows that the equation $R(u)u^{l-k}+s=0$ has no more than two roots $u$ in a neighborhood of $0$ if $l-k$ is even (and hence $\ge2$, by the Case 2.2 condition) and no more than one root $u$ in a neighborhood of $0$ if $l-k$ is odd. So,
$N_V(q+sp)\le k+2\le l=N_V(q)$ for some neighborhood $V$ of $v$ and all small enough $s>0$ if $l-k$ is even, and $N_V(q+sp)\le k+1\le l=N_V(q)$ for some neighborhood $V$ of $v$ and all small enough $s>0$ if $l-k$ is odd.

So, in all cases $N_V(q+sp)\le N_V(q)$ for some neighborhood $V$ of $v$ and all small enough $s>0$. Thus, (2) is proved. $\Box$