Proof:

We first define the associated maps $h_1:X\longrightarrow X\sqcup_fB$ and $h_2:X\longrightarrow X\sqcup_fB$. Let $\eta:X\sqcup B\longrightarrow X\sqcup_fB$ be the quotient map and $i_X:X\longrightarrow X\sqcup B$ and $i_B:B\longrightarrow X\sqcup B$ denote the inclusions. Then the associated maps $h_1$ and $h_2$ given by

$$h_1 = \eta\circ i_{X},\quad h_2 = \eta\circ i_{B}.\eqno(25.1)$$

For any $a \in A$ we have

$$h_1\circ i(a) = \eta(i_{X}(a)) = \eta(a),\quad h_2\circ f(a) = \eta(i_{B}(f(a))) = \eta(f(a))$$

Recalling the identifications we see that $h_1\circ i = h_2\circ f$. We now check the universal property. Suppose $Z$ is a topological space and $g_1:X\longrightarrow Z$, $g_2:B\longrightarrow Z$ are continuous maps such that

$$g_1\circ i = g_2\circ f \eqno(25.2)$$

Define the continuous map $\phi:X\sqcup B \longrightarrow Z$ as

$$\phi(x) = \left\{\begin{array}{lll} g_1(x) & & \mbox{if } x \in X \\ g_2(x) & & \mbox{if } x \in B. \\ \end{array} \right.$$

Condition (25.2) now shows that there is a unique map $\overline{\phi}:(X\sqcup_fB)/\sim\;\longrightarrow Z$ such that

$$\overline{\phi}\circ \eta = \phi. \eqno(25.3)$$

The universal property of the quotient implies that $\overline{\phi}$ is continuous. Equations (25.1)-(25.3) immediately give

$$\overline{\phi}\circ h_1 = g_1,\quad \overline{\phi}\circ h_2 = g_2.\eqno(25.4)$$

thereby completing the verification of the universal property.