Two Lie Group homomorphisms are equal if their induced Lie algebra homomorphisms are equal and $G$ is connected Let $f, g$ be two Lie Group homomorphisms from G to H (smooth group homomorphisms).
Let the induced Lie Algebra homomorphisms $Df(e)$ and $Dg(e)$ be equal. Then if $G$ is connected show that $f=g$.
I know that the set $S=\{x | f(x)=g(x)\}$ is closed. If I show it is open I am done. Now I know $e$ is in $S$. How do I show a nbd is also there?

Let $\varphi:G\to H$ a Lie group homomorphism and $d\varphi:\mathfrak{g}\to\mathfrak{h}$ the induced Lie algebra homomorphism, given by $(d\varphi X)_{e_H}=(d\varphi)_{e_G} X_{e_G}$.
Given $\varphi$, let $\Gamma_{\varphi}$ the graph of $\varphi$, namely $\Gamma_{\varphi}=\{(g,\varphi(g))\mid g\in G\}$. You can verify that $\Gamma_{\varphi}$ is an abstract subgroup of $G\times H$ and that $(\Gamma_{\varphi}, \iota)$ (where $\iota$ is the inclusion) is an embedding of $G\times H$ so it is a Lie subgroup of $G\times H$. 
What is the Lie algebra of $\Gamma_{\varphi}$ as a subalgebra of the Lie algebra of $G\times H$? Well, you can see that $\mathrm{Lie}(\Gamma_{\varphi})\cong \Gamma_{d\varphi}$, i.e. the graph of the homomorphism $d\varphi$.
Now the proof of your statement:
We have $\mathrm{Lie}(\Gamma_f)\cong \Gamma_{df}=\Gamma_{dg}\cong\mathrm{Lie}(\Gamma_g)$, and if $G$ is connected, then $\Gamma_f$ and $\Gamma_g$ are two connected subgroups which have the same Lie algebra. By uniqueness we must have $\Gamma_f=\Gamma_g$, i.e. $f(a)=g(a)\,\forall a\in G$.
