Proof:

It is clear that the homotopy is reflexive, and symmetry is left for student to verify. To prove transitivity let

$$F:\: [0,1] \times [0,1] \rightarrow X \;\;$$   and $$\;\; G:\: [0,1] \times [0,1] \rightarrow X$$

be homotopies between the pairs $\gamma_1,\gamma_2$ and $\gamma_2, \gamma_3$ respectively. Define $H : [0,1] \times [0,1] \rightarrow X$ by the prescription:

$$H(s, t) = \left\{\begin{array}{lll} F(2s, t) & & 0 \leq s \leq 1/2 \\ G(2s - 1) & & 1/2 \leq s \leq 1 \\ \end{array} \right.$$

Note that by gluing lemma $H$ is continuous. We need to check the conditions at endpoints.

$$H(s, 0) = \left\{\begin{array}{lll} F(2s, 0) = \gamma_1(0) = \gam... ... 0) = \gamma_2(0) = \gamma_3(0), & & 1/2 \leq s \leq 1 \\ \end{array} \right.$$

Likewise one verifies easily $H(s,1) = \gamma_1(1) = \gamma_3(1)$ for all $s\in [0,1]$. Finally we see that $H(0,t) = F(0,t) = \gamma_1(t)$ and $H(1,t) = G(1, t) = \gamma_3(t)$, which proves the result.