Notice first that from $f$ being a nontrivial solution there exists $x \in \RR$ such that $f(x) \neq 0$, so $f(x) = f(0+x) = f(0)f(x)$ and thus $f(0)=1$. Next, one can exploit the functional equation inductively to characterize $f$ when $x$ is an integer: $$ \begin{alignat*}2 f(x) &= f(1 + (x - 1)) = f(1)f(x-1) \stackrel{\text{IH}}{=} f(1)^x \qquad&&\text{if $x \in \ZZ^+$,}\\ f(x) &= f(-1 + (x + 1)) = f(-1)f(x+1) \stackrel{\text{IH}}{=} f(1)^x \qquad&&\text{if $x \in \ZZ^-$,} \end{alignat*} $$ where the last step in the second equation follows from the fact that $$ 1 = f(0) = f(x - x) = f(x)f(-x) \implies f(-x) = \frac1{f(x)}. $$ Clearly $f$ cannot have any roots, since assuming $f(\alpha) = 0$ for some $\alpha \in \RR$ yields $f(x) = f(\alpha+(x-\alpha)) = f(\alpha)f(x-\alpha)=0$, which contradicts the assumption that $f$ is not the zero function; likewise, it is always positive since $f(x)=f(x/2+x/2)=f(x/2)^2 > 0$. Hence, taking $p/q \in \QQ$ shows that $$ f(1)^p = f(p) = f\underbrace{\left(\frac pq+\cdots+\frac pq\right)}_{q \text{ times}} = f\left(\frac pq\right)^q \implies f\left(\frac pq\right) = f(1)^{p/q} $$ Finally, the definition of the continuity of $f$ states that, for any $x_0 \in \RR$, $\lim_{x\to x_0} f(x) = f(x_0)$; expanding the previous identity in terms of $\varepsilon$-$\delta$ and invoking the fact that $\QQ$ is dense in $\RR$ we see that for any $\varepsilon > 0$ there is a $p'/q' \in \QQ$ such that $|x_0 - p'/q'| < \delta$ and $|f(x_0) - f(1)^{p'/q'}| < \varepsilon$. Since $\delta$ gets arbitrarily small, it can be concluded that $f(x_0) = f(1)^{x_0}$ and thus $f(x) = f(1)^x$ for any $x \in \RR$, as it was to be proven.