real analysis - Continuous mapping of a compact metric space is uniformly continuous

Theorem: Let $f$ be a continuous mapping of a compact metric space $X$ into a metric space $Y$. Then $f$ is uniformly continuous on $X$.

Proof: Let $f$ is not uniformly continuous then for some $\varepsilon>0$ and arbitrary $\delta>0$ exists $x,y\in E$ and $d_X(x,y)<\delta$ but $d_Y(f(x),f(y))>\varepsilon.$ Taking $\delta=\frac{1}{n}$ then exists sequences $\{p_n\},\{q_n\}$ in $X$ such that $d_X(p_n,q_n)\to 0$ but $d_Y(f(p_n),f(q_n))>\varepsilon.$

Since $X$ is compact then $\{p_n\}$ and $\{q_n\}$ has limit point, namely $p$ and $q$. Then exists some subsequence $\{p_{n_k}\}$ that converges to $p$. Using triangle inequality we get that $\{q_{n_k}\}$ also converges to $p$.

Since $f$ is a continuous mapping then $f(p_{n_k})\to f(p)$ and $f(q_{n_k})\to f(p)$. Combine these statements we get $\varepsilon

Is this proof correct?