The real line with its additive group is a topological group?

Let $\mathbb R$ be the real line with its additive group structure and euclidean topology. We want to prove that the real line is indeed a topological group. Let $i:\mathbb R \to \mathbb R$, defined by $i(x)=-x$ and $m:\mathbb R \times \mathbb R \to \mathbb R$ defined by $m(x,y)= x+y$. we have to prove the following: (a) $i$ is continuous (b) $m$ is continuous.

(a) By a real analysis argument we know that $i$ is continuous, because $i$ is a polynomial function with real coefficients.

(b) We know that the projection $\Pi_1:X\times Y \to X$, $\Pi_1(x,y)=x$, where $X$ and $Y$ are topological spaces, is always continuous because for any open subset $U$ of X, we have $\Pi^{-1}(U)=U\times Y$ a open subset of $X \times Y$.

With the same argument the projection $\Pi_2:X\times Y \to Y$, $\Pi_2(x,y)=y$, where $X$ and $Y$ are topological spaces, is also continuous.

So by a real analysis argument which claims that the sum of continuous functions is continuous and as we know $m =\Pi_1 +\Pi_2$, then m is continuous.

Addition is continuous because of the familiar 'addition law for limits':

$$\lim x_n=x \text{ and }\lim y_n=y\ \Rightarrow \lim (x_n+y_n)=x+y.$$

Let $g_1,g_2 \in \mathbb{R}$ and $\epsilon>0$.

Then for each $(r_1,r_2)\in (g_1-\frac{\epsilon}{2},g_1+\frac{\epsilon}{2}) \times (g_2-\frac{\epsilon}{2},g_2+\frac{\epsilon}{2})$ we have that $$m(r_1,r_2)=r_1+r_2 \in (m(g_1,g_2)-\epsilon,m(g_1,g_2)+\epsilon) \ .$$

So if the definition for continuous functions you have is:

$f:A\longrightarrow B$ is continuous at $a \in A$ if for every open neighborhood $U$ of $f(a) \in B$ exist an open neighborhood $V$ of $a \in A$ such that $$\forall x \in V \; \Rightarrow\; f(x) \in U \ .$$

then $m$ is continuous.