As in other two of my questions, which are already answered by myself, I am treating exponential function again. Now, from the perspective of continuity only. These means, I can not use any single argument about derivatives. I will show the layout of this project and at the end I will set up my problem.
Let $f:\mathbb{R}\longrightarrow\mathbb{R}$, such that $f(x+y)=f(x)f(y)$.
Theorem. If $f$ is continuous in $c$ then $f$ is continuous, $f(0)=1$, $f(rx)=(f(x))^{r},\,\forall r\in\mathbb{Q}$ and if $f(x)=0$ for some $x\in\mathbb{R}$ then $f=0$. Moreover, $f$ is uniformly continuous if it is continuous.
Proof. We have $f(0)=f(0+0)=f(0)f(0)=(f(0))^{2}$, then $f(0)=1$ o $f(0)=0$. Also $f(nx)=(f(x))^{n}$, with $n\in\mathbb{N}$, since $f(x)=f(x)$, $f(2x)=f(x+x)=f(x)f(x)=(f(x))^{2}$ and if $f(nx)=(f(x))^{n}$ entonces $f((n+1)x)=f(nx+x)=f(nx)f(x)=(f(x))^{n}f(x)=(f(x))^{n+1}$. Also $f(0)=f(x+(-x))=f(x)f(-x)$ then $f(-x)=\frac{1}{f(x)}=(f(x))^{-1}$, if $f(0)=1$ and $f(x)=0\lor f(-x)=0$ if $f(0)=0$. Finally $(f(\frac{n}{m}x))^{m}=f(nx)=(f(x))^{n}$ therefore $f(\frac{n}{m}x)=(f(x))^{\frac{n}{m}}$, that is, for every rational $r$ it is true that $f(rx)=(f(x))^{\frac{n}{m}}$.
We can suspect that $f(0)=0\longrightarrow \forall x,\,f(x)=0$. Readily $f(x)=f(x+0)=f(x)f(0)=0f(x)=0$. Then if $f(0)=0$ then $f=0$ the constant function such that $x\mapsto 0$. Moreover, if $f(x)=0$ for some $x\neq 0$ then $f(y)=f(y+x-x)=f(y-x)f(x)=0$. That is, if $f(x)=0$ for some $x$ then $f$ vanishes for every $x$.
Thus, if $f=0$ we know is continuous since constant functions are continuous (and in fact uniformly continuous). Therefore we need $f(0)=1$ and then we can be sure that $f(x)\neq 0$ for every $x$.
We know that $f$ is continuous on $c$: $\forall\epsilon>0\,\exists\delta_{\epsilon,c}>0\mid\forall x\in B_{c,\delta_{\epsilon,c}}\longrightarrow f(x)\in B_{f(c),\epsilon}$.
On the other hand, if we fix $x_{0}\in\mathbb{R}$, then $f((x+x_{0})-c)=\frac{f(x)}{f(c)}f(x_{0})$ y $f((x+x_{0})-c)-f(x_{0})=\left(\frac{f(x)}{f(c)}-1\right)f(x_{0})=\frac{f(c)}{f(c)}\left(\frac{f(x)}{f(c)}-1\right)f(x_{0})=\frac{f(x_{0})}{f(c)}\left(f(x)-f(c)\right)$.
Given $x_{0}$ we can do the following mapping $\phi:\mathbb{R}\longrightarrow\mathbb{R}$, $\phi(x)=x+(x_{0}-c)=id_{V}(x)+d$, and evidently $\psi(x)=x+(c-x_{0})$. Then $(\phi\circ\psi)(x)=\phi(\psi(x))=x+(c-x_{0})+(x_{0}-c)=x=id_{\mathbb{R}}(x)$ and $(\psi\circ\phi)(x)=\psi(\phi(x))=x+(x_{0}-c)+(c-x_{0})=x=id_{\mathbb{R}}(x)$. Then $\phi\circ\psi=\psi\circ\phi=id_{V}$, and therefore $\psi=\phi^{-1}$. Moreover, since identity is continuous and constants are also continuous, $\phi$ is continuous and a homeomorphism.
The former mapping takes, homeomorphically, the ball $B_{c,\delta_{\epsilon,c}}$ onto $B_{x_{0},\delta_{\epsilon,c}}$, and since it is only a translation there is no scale change. This allows us to do the following.
We have $\left\vert x-c\right\vert<\delta_{\epsilon, c}$, and applying the homeomorphism $\left\vert(x+(x_{0}-c))-x_{0}\right\vert<\delta_{\epsilon, c}$ and since it is a bijection every point in the ball centered at $x_{0}$ can be written as $x+(x_{0}-c)$. Then $\left\vert f(y)-f(x_{0})\right\vert$ with $y\in B_{x_{0},\delta_{\epsilon,c}}$, using $\phi$ on $x\in B_{c,\delta_{\epsilon,c}}$ it follows $\left\vert f(x+(x_{0}-c))-f(x_{0})\right\vert=\left\vert f(x)+f(x_{0})-f(c)-f(x_{0})\right\vert=\left\vert f(x)-f(c)\right\vert<\epsilon$. Therefore it implies that: $\forall\epsilon>0,\,x_{0}\in\mathbb{R}\,\exists\delta_{\epsilon,x_{0}}>0\mid\forall x\in B_{x_{0},\delta_{\epsilon,x_{0}}}\longrightarrow f(x)\in B_{f(x_{0}),\epsilon}$ and we found that $\delta_{\epsilon,c}=\delta_{\epsilon,x_{0}}$ forall $x_{0}\in\mathbb{R}$, then $\delta_{\epsilon,x_{0}}=\delta_{\epsilon}$.
This confirms that $f$ is continuous, and uniformly continuous.
Let us notice that functional equation gives one of the properties of exponential functions, when this does not vanish identically. From previous theorem we can readily prove that
Theorem. Let $f:\mathbb{R}\longrightarrow\mathbb{R}$ such that $f(x+y)=f(x)f(y)$ and $f\neq 0$. If $f(1)=a$, then $a>0$ and $f(r)=a^{r}$ for every $r\in\mathbb{Q}$.
Proof. By former theorem, and since $r=r\cdot 1$, then $f(r)=f(r\cdot1)=\left(f(1)\right)^{r}$. Thus, if $f(1)=a$ then $f(r)=a^{r}$.
Additionally, suppose that $a<0$, since if equality is allowed, then $f$ would vanish identically. Notice that $f\left(\frac{1}{2}\right)=a^{\frac{1}{2}}$. That is $a^{\frac{1}{2}}=\xi\in\mathbb{R}$ such that $\xi^{2}=a$. But $x^{2}\geq 0$ for every $x\in\mathbb{R}$. Then we are telling that $a\geq 0$, but we have supposed that $a<0$. Thus, there is no other choice but $a>0$.
Lemma. Let $f:\mathbb{R}\longrightarrow\mathbb{R}$ such that $f(x+y)=f(x)f(y)$ and $f\neq 0$. If $f(x)>1$ for $0<x<\delta$ with $\delta$ some positive real number, then $f(y)>0$ for every $y\in\mathbb{R}$. Moreover $f$ is strictly monotonic increasing.
Proof. It suffices to prove this for $y\geq 0$ since every negative number could be written as $-y$ for some positive $y$ and $f(-y)=\frac{1}{f(y)}$. It is clear for $y=0$ since $f(0)=1>0$ HOW CAN I PROVE THIS FOR $y>0$? FOR SHOWING THAT $f$ IS STRICTLY MONOTONIC INCREASING FUNCTION FROM POSITIVITY I HAVE NO IDEA!
CAN ANYONE SHARE SOME HINTS OR SOLUTIONS?
We look at part of your problem. We want to show that if $0\lt a\lt b$ then $f(a)\lt f(b)$.
Note that $f(b)=f(a+(b-a))=f(a)f(b-a)$. So we need only show that $f(b-a)\gt 1$.
So we want to show that if $c\gt 0$ then $f(c)\gt 1$.
There is a number $\epsilon$ in the interval $(0,\delta)$, and a positive integer $n$, such that $n\epsilon=c$.
By induction from $f(x+y)=f(x)f(y)$, we can prove that $f(n\epsilon)=(f(\epsilon))^n$. Since $f(\epsilon)\gt 1$, it follows that $(f(\epsilon))^n\gt 1$. So $f(c)\gt 1$, and we are finished.