Is this proof of uniqueness of the limit correct? I've tried to show the following: let $(M,d_M)$ and $(N,d_N)$ be metric spaces and $f : M \to N$. If $a \in M$ and $\lim_{p\to a}f(p)$ exists, then it is unique. 
I'm a little unsure if the proof is correct, mainly the use of the squeeze theorem in the end. Is this proof a good one?
My proof was as follows: suppose there were two limits, $L_1$ and $L_2$, then by the definition, given $\epsilon/2 > 0$ there would be $\delta_1 > 0$ and $\delta_2 > 0$ such that if we take $\delta = \min\{\delta_1,\delta_2\}$, then this means that $d_M(p,a) < \delta$ implies $d_N(f(p),L_1)<\epsilon/2$ and at the same time $d_N(f(p),L_2)<\epsilon/2$. 
Now, summing these two inequalities, we have $d_N(f(p),L_1)+d_N(f(p),L_2)<\epsilon$, but $d_N(f(p),L_1)=d(L_1,f(p))$ and thus $d_N(L_1,f(p))+d_N(f(p),L_2)<\epsilon$, now since we have triangle inequality available, we have $d_N(L_1,L_2)\leq d_N(L_1, f(p))+d_N(f(p),L_2)<\epsilon$. Since $\epsilon>0$ and since $d_N$ is metric, we now have that $-\epsilon < d_N(L_1,L_2)<\epsilon$ and then if we define $\phi_1,\phi,\phi_2 : \mathbb{R} \to \mathbb{R}$ by $\phi_1(\epsilon)=-\epsilon$, $\phi(\epsilon)=d_N(L_1,L_2)$ and $\phi_2(\epsilon)=\epsilon$ we found out that in reality we have:
$$\phi_1(\epsilon) < \phi(\epsilon) < \phi_2(\epsilon)$$
Now taking the limit in $\mathbb{R}$ when $\epsilon$ goes to zero, $\phi_1(\epsilon)$ and $\phi_2(\epsilon)$ goes to zero, and so the limit of $\phi(\epsilon)$ with $\epsilon$ going to zero, must be zero. But since $\phi$ independs on $\epsilon$ this implies that $\phi = 0$ and so $d_N(L_1,L_2) = 0$ implying by properties of metric that $L_1 = L_2$.
I felt this proof too simple, so I'm almost certain there is some little detail I forgot about. Is it correct?
Thanks very much in advance!
 A: Your proof is essentially correct, but the style is not the best.
When you have shown that $\;d_N(L_1,L_2)<\epsilon$, you are done because $\;d_N(L_1,L_2) \ge 0$.
Remember the proposition:  

If $\; 0 \le a < \epsilon$ for every $\epsilon>0$, then $a=0$.

You see this in the proof of a Master:
J.Dieudonné Foundations of Modern Analysis (1960), prop. (3.13.3) pag. 47.
A: Your proof seems correct. You could proceed quicker though.
Suppose not. Then there are two limits $L_{1}$ and $L_{2}$. Take $0<\epsilon=d_{N}(L_{1},L_{2})$. By the definition of limit there is $\delta_{1}$ and $\delta_{2}$ such that if $d_{M}(p,a)<\delta_{1}$ then $d_{N}(f(p),L_{1})<\frac{\epsilon}{100}$ and if $d_{M}(p,a)<\delta_{2}$ then $d_{N}(f(p),L_{2})<\frac{\epsilon}{100}$. Let $\delta=\min\{\delta_{1},\delta_{2}\}$. Then if $d_{M}(p,a)<\delta$ then $d_{N}(L_{1},L_{2})\le d_{N}(L_{1},f(p))+d_{N}(f(p),L_{2})<\frac{\epsilon}{100}+\frac{\epsilon}{100}=\frac{\epsilon}{50}=\frac{1}{50}d_{N}(L_{1},L_{2})$. So $49d_{N}(L_{1},L_{2})<0$ which is impossible.
