Let $X,Y,Z \in \mathbf{Top}_*$ be pointed spaces with basepoints $x_0,y_0$ and $z_0$. Then the wedge-sum $X\vee Y = X \sqcup Y / (x_0 \sim y_0)$ is a coproduct of $X$ and $Y$. Especially given pointed maps $f:X \to Z$ and $g:Y \to Z$ the map $(f,g)$ should be continous where $$ (f,g)([p,\delta]) = \begin{cases} f(p) & \delta = 0 \\ g(p) & \delta = 1 \end{cases} $$ In order to prove continuity let $U \subset Z$ be open. Then $$ (f,g)^{-1}(U) = i_X(f^{-1}(U)) \cup i_Y(g^{-1}(U)) $$ where $i_X:X \to X \vee Y: x \mapsto [x,0]$ and similar $i_Y$. Clearly, we are done if $i_X$ and $i_Y$ are open maps.
In order to see that $i_X$ is open let $V \subset X$ be open. Let $p:X \sqcup Y \to X \vee Y: (p,\delta) \mapsto [p,\delta]$. We want that $i_X(U)$ is open in $X \vee Y$, which is true if $p^{-1}(i_X(U))$ is open in $ X \sqcup Y$. If now $x_0 \in U$ then $p^{-1}(i_X(U)) = i_X(U) \cup \{(y_0,1)\}$. Let $\hat i_X: X \mapsto X \sqcup Y: x \mapsto (x,0)$. We are done if $$ \hat i_X^{-1}(i_X(U) \cup \{(y_0,1)\}) = U $$ and $$ \hat i_Y^{-1}(i_X(U) \cup \{(y_0,1)\}) = \{y_0\} $$ are open in $X$ resp. $Y$.
By $\{y_0\}$ does not have to be open. Apparently, this approach does not work.
How can I prove continuity ?
Let $q$ be the quotient map from $X \sqcup Y$ onto $X\vee Y$, where the $q$ just identifies $x_0$ and $y_0$ and no other points. Then $(f,g)^{-1}[U]$ is open in $X\vee Y$ iff $q^{-1}[(f,g)^{-1}[U]]$ is open in $X \sqcup Y$.
And letting $j_X: X \rightarrow X \sqcup Y$ and $j_Y: Y \rightarrow X \sqcup Y$ be the standard maps $j_X(x) = (x,0), j_Y(y) = (y,1)$, we also know by the sum topology properties (it's the final topology w.r.t. $j_X$ and $j_Y$) that a set $O$ in $X \sqcup Y$ is open iff both $(j_X)^{-1}[O]$ is open in $X$ and $(j_Y)^{-1}[O]$ is open in $Y$.
So $q^{-1}[(f,g)^{-1}[U]]$ is open in $X\sqcup Y$ iff $(j_X)^{-1}[q^{-1}[(f,g)^{-1}[U]]]$ is open in $X$ and $(j_Y)^{-1}[q^{-1}[(f,g)^{-1}[U]]]$ is open in $Y$.
Unwrapping the definitions will give that the former is just $f^{-1}[U]$ and the latter is $g^{-1}[U]$, and so you are done.