Wazzup!
How do I get around proving (by an indirect argument) that a continuos mapping f: X -> Y between metric-spaces (X,d_X) and (Y,d_y) is uniformly continuous in every sequentially compact subset K of X.
My suggestion is to suppose that $\displaystyle f$ is not uniformly continuous. Then, there exists $\displaystyle \varepsilon> 0$ such that for every $\displaystyle \delta>0$ there exist $\displaystyle x,y\in X$ such that $\displaystyle d(x,y)<\delta$ but $\displaystyle d(f(x),f(y))\geq \varepsilon$.
Try to construct a sequence that behaves badly. Keep in mind that the space is sequentially compact.
You first need to construct some kind of sequence out of the failure of uniform continuity. For example, for $\displaystyle \delta=1$ we can find points $\displaystyle x_1,y_1\in X$ such that $\displaystyle d(x_1,y_1)<1$ but $\displaystyle d(f(x_1),f(y_1))\geq \varepsilon$. For $\displaystyle \delta=1/2$, we can find $\displaystyle x_2,y_2\in X$ such that $\displaystyle d(x_2,y_2)<1/2$ but $\displaystyle d(f(x_2),f(y_2))\geq \varepsilon$. Continuing the process, we can find $\displaystyle \delta=1/n$ such that $\displaystyle d(x_n,y_n)<1/n$ but $\displaystyle d(f(x_n),f(y_n))\geq \varepsilon$.
This gives you two sequences $\displaystyle \{x_n\}$ and $\displaystyle \{y_n\}$ such that $\displaystyle d(x_n,y_n)<1/n$ and $\displaystyle d(f(x_n),f(y_n))\geq \varepsilon$ for every $\displaystyle n$. Use sequential compactness here to relate them to the continuity of $\displaystyle f$.
Note: This is more high-powered for than roninpro's suggestion. If it's gobbeldygook to you just ignore it.
It's trivial that for metric spaces sequential compactness and compactness are synonymous. So, let $\displaystyle \varepsilon>0$ be arbitrary and cover $\displaystyle f(K)$ with the open balls $\displaystyle \left\{B_{\varepsilon}(f(x))\right\}_{x\in K}$. Clearly then we have that $\displaystyle \left\{f^{-1}\left(B_{\varepsilon}(f(x))\right)\right\}_{x\in K}$ is an open cover for $\displaystyle K$. Then, since $\displaystyle K$ is compact we know that the open cover $\displaystyle \left\{f^{-1}\left(B_{\varepsilon}\left(f(x)\right)\right)\ri ght\}_{x\in K}$ admits a Lebesgue number $\displaystyle \delta$. Note then that if $\displaystyle d_X(x,y)<\delta$ then we have that $\displaystyle \text{diam}\{x,y\}<\delta$ (this last part was pedantic, but just in case) and so by definition we have there is some $\displaystyle f^{-1}\left(B_{\varepsilon}(f(x_0))\right)$ such that $\displaystyle x,y\in f^{-1}\left(B_{\varepsilon}(f(x_0))\right)$ and so $\displaystyle f(x),f(y)\in B_{\varepsilon}(f(x))$ and thus $\displaystyle d_Y(f(x),f(y))<2\varepsilon$.