Convolution of even functions is even
Just do a change of variables:

\begin{align}
z(t) &amp;= x\star y \big|_{t}\\
&amp;=\int_{\tau=-\infty}^{\tau = \infty} x(\tau)y(t-\tau) \,\mathrm d\tau
&amp;\scriptstyle{\text{Set}~\tau=-\lambda, \mathrm d\tau=-\mathrm d\lambda} \tag{1}\\
&amp;=-\int_{-\lambda=-\infty}^{-\lambda = \infty} x(-\lambda)y(t+\lambda) \,\mathrm d\lambda &amp;\scriptstyle{\text{Now remember that}~ x~\text{and}~y
~\text{are even functions}}\\ 
&amp;=\int_{\lambda=-\infty}^{\lambda=\infty} x(\lambda)y((-t)-\lambda) \,\mathrm d\lambda &amp;\scriptstyle{\text{Same as (1) except with }\lambda~\text{instead of }\tau ~\text{and with}~-t}\\
&amp;= z(-t)\end{align} 
with nary a mention of LTI systems. 

For a more DSPish proof, consider that the Fourier transform of a real-valued even function of time $t$ is a real-valued even function of frequency $f$, and thus if $X(f)$ and $Y(f)$ both are real-valued even functions of $f$, then so is $Z(f) =X(f)Y(f)$ also a real-valued even function of $f$, implying in turn that $z = x\star y$ is a real-valued even function of $t$ whenever $x(t)$ and $y(t)$ are real-valued even functions of $t$.